CBSE Class 11 Mathematics Trigonometric Equations Worksheet Set 01

INTEGER QUESTIONS

Question. If \( x \in [0, 2\pi] \) for which \( 2\cos x \leq |\sqrt{1+\sin 2x} - \sqrt{1-\sin 2x}| \leq \sqrt{2} \) has solution set \( x \in \left[ \frac{\lambda\pi}{4}, \frac{\mu\pi}{4} \right] \) then \( \mu - \lambda \) is
Answer: Let \( y = |\sqrt{1+\sin 2x} - \sqrt{1-\sin 2x}| \)

\( \implies \) \( y^2 = 2 - 2|\cos 2x| \)
If \( x \in \left[ 0, \frac{\pi}{4} \right] \) or \( \left[ \frac{3\pi}{4}, \frac{5\pi}{4} \right] \) or \( \left[ \frac{7\pi}{4}, 2\pi \right] \)
\( \cos 2x \) is non - negative
so, \( y^2 = 2 - 2\cos 2x = 4\sin^2 x \)
\( y = 2|\sin x| \)

\( \implies \) \( \cos x \leq |\sin x| \)
Except for \( x \) in \( \left[ 0, \frac{\pi}{4} \right] \) and \( \left[ \frac{7\pi}{4}, 2\pi \right] \)
so, that leaves \( \left[ \frac{3\pi}{4}, \frac{5\pi}{4} \right] \)
In which we certainly have \( \sin x \leq \frac{1}{\sqrt{2}} \)
If \( x \in \left( \frac{\pi}{4}, \frac{3\pi}{4} \right) \) or \( \left( \frac{5\pi}{4}, \frac{7\pi}{4} \right) \) then
\( \cos 2x \) is negative, so
\( y^2 = 2 + 2\cos 2x = 4\cos^2 x \)

\( \implies \) \( y = 2|\cos x| \)
so, the first inequality certainly holds the second also holds
Thus, solution set is \( x \in \left[ \frac{\pi}{4}, \frac{7\pi}{4} \right] \)

\( \implies \) \( \mu - \lambda = 6 \)

Question. The number of integral values of \( \alpha \) such that \( \sin x \cos 3x - \alpha \cos x \sin 3x = 0 \) does not have any real root other than \( (2n+1)\frac{\pi}{2}, n \in I \) for any real value of \( x \), is
Answer: \( \sin x \cos 3x - \alpha \cos x \sin 3x = 0 \) ....(i)

\( \implies \) \( \alpha = \frac{\sin x \cos 3x}{\cos x \sin 3x} \)
\( \alpha = \frac{\tan x}{\tan 3x} \)
\( \alpha = \frac{\tan x (1 - 3\tan^2 x)}{3\tan x - \tan^3 x} \)
\( \alpha = \frac{1 - 3\tan^2 x}{3 - \tan^2 x} \) ....(ii)
For real value of x, RHS of 2nd never lies between \( \left[ \frac{1}{3}, 3 \right) \)
\( \implies \) Number of integral points {1,2,3}
\( \implies \) 3 points

Question. If \( 4\cos 36^\circ + \cot\left( 7\frac{1}{2}^\circ \right) = \sqrt{n_1} + \sqrt{n_2} + \sqrt{n_3} + \sqrt{n_4} + \sqrt{n_5} + \sqrt{n_6} \) then the product of the digits in \( \sum_{i=1}^6 n_i^2 = \)
Answer: \( \because \cot\left( 7\frac{1}{2}^\circ \right) = \frac{1 + \cos 15^\circ}{\sin 15^\circ} \)
\( = \frac{1 + \frac{\sqrt{3}+1}{2\sqrt{2}}}{\frac{\sqrt{3}-1}{2\sqrt{2}}} = \frac{2\sqrt{2} + \sqrt{3} + 1}{\sqrt{3} - 1} \)
\( = \frac{(2\sqrt{2} + \sqrt{3} + 1)(\sqrt{3} + 1)}{2} \)
\( = \frac{2\sqrt{6} + 2\sqrt{2} + 3 + \sqrt{3} + \sqrt{3} + 1}{2} \)
\( = \sqrt{6} + \sqrt{2} + 2 + \sqrt{3} \)
\( = \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6} \)
and
\( 4\cos 36^\circ = 4\left( \frac{\sqrt{5}+1}{4} \right) = \sqrt{5} + 1 = \sqrt{5} + \sqrt{1} \)
Hence, \( 4\cos 36^\circ + \cot\left( 7\frac{1}{2}^\circ \right) \)
\( = \sqrt{1} + \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{5} + \sqrt{6} \)
\( \therefore n_1 = 1, n_2 = 2, n_3 = 3, n_4 = 4 \)
\( n_5 = 5 \) and \( n_6 = 6 \)
\( \therefore \sum_{i=1}^6 n_i^2 = n_1^2 + n_2^2 + n_3^2 + n_4^2 + n_5^2 + n_6^2 \)
\( = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 \)
\( = 91 \)

Question. If \( \tan^2\left(\frac{\pi}{16}\right) + \tan^2\left(\frac{2\pi}{16}\right) + \tan^2\left(\frac{3\pi}{16}\right) + ... + \tan^2\left(\frac{7\pi}{16}\right) = \lambda \) and if \( x^y + y^x = \lambda (x, y \in Z^+) \) then the sum of the digits in \( (x+y) \) is
Answer: \( \lambda = \left\{ \tan^2\left(\frac{\pi}{16}\right) + \tan^2\left(\frac{7\pi}{16}\right) \right\} \)
\( + \left\{ \tan^2\left(\frac{2\pi}{16}\right) + \tan^2\left(\frac{6\pi}{16}\right) \right\} \)
\( + \left\{ \tan^2\left(\frac{3\pi}{16}\right) + \tan^2\left(\frac{5\pi}{16}\right) \right\} + \tan^2\left(\frac{4\pi}{16}\right) \)
\( = \tan^2\left(\frac{\pi}{16}\right) + \cot^2\left(\frac{\pi}{2} - \frac{7\pi}{16}\right) \)
\( + \left\{ \tan^2\left(\frac{2\pi}{16}\right) + \cot^2\left(\frac{\pi}{2} - \frac{6\pi}{16}\right) \right\} \)
\( + \left\{ \tan^2\left(\frac{3\pi}{16}\right) + \cot^2\left(\frac{\pi}{2} - \frac{5\pi}{16}\right) \right\} + 1 \)
\( = \left\{ \tan^2\left(\frac{\pi}{16}\right) + \cot^2\left(\frac{\pi}{16}\right) \right\} \)
\( + \left\{ \tan^2\left(\frac{2\pi}{16}\right) + \cot^2\left(\frac{2\pi}{16}\right) \right\} \)
\( + \left\{ \tan^2\left(\frac{3\pi}{16}\right) + \cot^2\left(\frac{3\pi}{16}\right) \right\} + 1 \)
\( = \left\{ \tan\left(\frac{\pi}{16}\right) + \cot\left(\frac{\pi}{16}\right) \right\}^2 + \left\{ \tan\left(\frac{2\pi}{16}\right) + \cot\left(\frac{2\pi}{16}\right) \right\}^2 \)
\( + \left\{ \tan\left(\frac{3\pi}{11}\right) + \cot\left(\frac{3\pi}{16}\right) \right\}^2 - 2 - 2 - 2 + 1 \)
\( = \frac{1}{\left\{ \sin\left(\frac{\pi}{16}\right)\cos\left(\frac{\pi}{16}\right) \right\}^2} + \frac{1}{\left\{ \sin\left(\frac{2\pi}{16}\right)\cos\left(\frac{2\pi}{16}\right) \right\}^2} \)
\( + \frac{1}{\left\{ \sin\left(\frac{3\pi}{16}\right)\cos\left(\frac{3\pi}{16}\right) \right\}^2} - 5 \)
\( = \frac{4}{\sin^2\left(\frac{\pi}{8}\right)} + \frac{4}{\sin^2\left(\frac{\pi}{4}\right)} + \frac{4}{\sin^2\left(\frac{3\pi}{8}\right)} - 5 \)
\( = 4\left\{ \frac{1}{\sin^2\left(\frac{\pi}{8}\right)} + \frac{1}{\sin^2\left(\frac{3\pi}{8}\right)} \right\} + 4.2 - 5 \)
\( = 4\left\{ \frac{1}{\sin^2\left(\frac{\pi}{8}\right)} + \frac{1}{\cos^2\left(\frac{\pi}{8}\right)} \right\} + 3 \)
\( = \frac{4}{\left\{ \sin\left(\frac{\pi}{8}\right)\cos\left(\frac{\pi}{8}\right) \right\}^2} + 3 \)
\( = \frac{16}{\left\{ \sin\left(\frac{\pi}{4}\right) \right\}^2} + 3 \)
\( \therefore \lambda = 35 \)
Then, \( x^y + y^x = 35 \)
\( \implies \) \( x = 34, y = 1 \)
or x=1, y=34

Question. Suppose \( \cos\frac{2\pi}{7} + \cos\frac{4\pi}{7} + \cos\frac{6\pi}{7} = -\frac{1}{2} \) and \( \cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{6\pi}{7} = \frac{1}{8} \), then the numerical value of \( \csc^2\frac{\pi}{7} + \csc^2\frac{2\pi}{7} + \csc^2\frac{3\pi}{7} \) must be
Answer:  \( \csc^2\frac{\pi}{7} + \csc^2\frac{2\pi}{7} + \csc^2\frac{3\pi}{7} \)
\( = \frac{1}{\sin^2\frac{\pi}{7}} + \frac{1}{\sin^2\frac{2\pi}{7}} + \frac{1}{\sin^2\frac{3\pi}{7}} \)
\( = \frac{2}{1 - \cos\frac{2\pi}{7}} + \frac{2}{1 - \cos\frac{4\pi}{7}} + \frac{2}{1 - \cos\frac{6\pi}{7}} \)
Now denominator (after taking LCM)
\( = \left( 1 - \cos\frac{2\pi}{7} \right)\left( 1 - \cos\frac{4\pi}{7} \right)\left( 1 - \cos\frac{6\pi}{7} \right) \)
\( = 1 - \left( \cos\frac{2\pi}{7} + \cos\frac{4\pi}{7} + \cos\frac{6\pi}{7} \right) \)
\( + \left( \cos\frac{2\pi}{7}\cos\frac{4\pi}{7} + \cos\frac{4\pi}{7}\cos\frac{6\pi}{7} + \cos\frac{6\pi}{7}\cos\frac{2\pi}{7} \right) \)
\( - \cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{6\pi}{7} \)
The second and last terms are known in the light of given relations.
The third term
\( = \frac{1}{2} \left[ \cos\frac{6\pi}{7} + \cos\frac{2\pi}{7} + \cos\frac{10\pi}{7} + \cos\frac{2\pi}{7} + \cos\frac{8\pi}{7} + \cos\frac{4\pi}{7} \right] \)
\( = \cos\frac{2\pi}{7} + \cos\frac{4\pi}{7} + \cos\frac{6\pi}{7} = -\frac{1}{2} \)
This denomirator \( = 1 - \left( -\frac{1}{2} \right) - \frac{1}{2} + \frac{1}{8} = \frac{7}{8} \)
The numerator
\( = 2 \left[ 3 - 2\left( \cos\frac{2\pi}{7} + \cos\frac{4\pi}{7} + \cos\frac{6\pi}{7} \right) \right. \)
\( \left. + \cos\frac{2\pi}{7}\cos\frac{4\pi}{7} + \cos\frac{4\pi}{7}\cos\frac{6\pi}{7} + \cos\frac{6\pi}{7}\cos\frac{2\pi}{7} \right] \)
\( = 2 \left[ 3 - 2\left( -\frac{1}{2} \right) - \frac{1}{2} \right] = 7 \). Whence the given expression \( = \frac{7}{7/8} = 8 \)

Question. If \( \tan x + \tan 2x + \tan 3x = \tan x \tan 2x \tan 3x \) then \( |\sin 3x + \cos 3x| = \)
Answer: \( \tan 3x = 0 \)
\( \implies \) \( x = \frac{n\pi}{3}, n \in Z \)

Question. \( 16\left(\cos\theta - \cos\frac{\pi}{8}\right)\left(\cos\theta - \cos\frac{3\pi}{8}\right)\left(\cos\theta - \cos\frac{5\pi}{8}\right)\left(\cos\theta - \cos\frac{7\pi}{8}\right) = \lambda\cos 4\theta \) then the value of \( \lambda \) is
Answer: LHS = \( 16\left(\cos\theta - \cos\frac{\pi}{8}\right)\left(\cos\theta - \cos\frac{3\pi}{8}\right) \times \left(\cos\theta - \cos\frac{5\pi}{8}\right)\left(\cos\theta - \cos\frac{7\pi}{8}\right) \)
\( = 16\left(\cos^2\theta - \cos^2\frac{\pi}{8}\right)\left(\cos^2\theta - \cos^2\frac{3\pi}{8}\right) \)
\( = 16\left(\cos^4\theta - \cos^2\theta + \sin^2\frac{\pi}{8}\cos^2\frac{\pi}{8}\right) \)
\( = 16\left(\cos^4\theta - \cos^2\theta + \frac{1}{8}\right) \)
\( = 16\left(-\cos^2\theta\sin^2\theta + \frac{1}{8}\right) = 16\left(\frac{-\sin^2 2\theta}{4} + \frac{1}{8}\right) \)
\( 16\left(\frac{1 - 2\sin^2 2\theta}{8}\right) = 16\left(\frac{\cos^2 2\theta - \sin^2 2\theta}{8}\right) \)
\( = \frac{16\cos 4\theta}{8} = 2\cos 4\theta \)

Question. Given that \( f(n\theta) = \frac{2\sin 2\theta}{\cos 2\theta - \cos 4n\theta} \) and \( f(\theta) + f(2\theta) + f(3\theta) + \dots + f(n\theta) = \frac{\sin \lambda\theta}{\sin \theta \sin \mu\theta} \) then the value of \( \mu - \lambda \), is
Answer: \( f(n\theta) = \frac{2\sin 2\theta}{\cos 2\theta - \cos 4n\theta} \)
\( = \frac{2\sin 2\theta}{2\sin(2n+1)\theta\sin(2n-1)\theta} \)
\( = \frac{\sin((2n+1)\theta - (2n-1)\theta)}{\sin(2n+1)\theta\sin(2n-1)\theta} \)
\( = \frac{\sin(2n+1)\theta\cos(2n-1)\theta - \cos(2n+1)\theta\sin(2n-1)\theta}{\sin(2n+1)\theta\sin(2n-1)\theta} \)
\( = \cot(2n-1)\theta - \cot(2n+1)\theta \)

Question. If \( 4\sin 27^\circ = \sqrt{\alpha} + \sqrt{\beta} \) then the sum of the digits in \( (\alpha+\beta-\alpha\beta+2)^4 \)
Answer: \( \because (\cos 27^\circ + \sin 27^\circ)^2 = 1 + \sin 54^\circ = 1 + \cos 36^\circ \)

\( \implies \) \( \cos 27^\circ + \sin 27^\circ = \sqrt{(1 + \cos 36^\circ)} \)
Also, \( \cos 27^\circ - \sin 27^\circ = \sqrt{(1 - \cos 36^\circ)} \)
\( (\because \cos 27^\circ > \sin 27^\circ) \)
\( \therefore 2\sin 27^\circ = \sqrt{(1 + \cos 36^\circ)} - \sqrt{(1 - \cos 36^\circ)} \)
\( = \sqrt{\left(1 + \frac{\sqrt{5}+1}{4}\right)} - \sqrt{\left(1 - \frac{\sqrt{5}+1}{4}\right)} \)
\( \therefore 4\sin 27^\circ = \sqrt{(5 + \sqrt{5})} - \sqrt{(3 - \sqrt{5})} \)
on comparing, we get
\( \alpha = 5 + \sqrt{5}, \beta = 3 - \sqrt{5} \)
\( \therefore \alpha + \beta = 8, \alpha\beta = 10 - 2\sqrt{5} \)
\( \alpha + \beta - \alpha\beta + 2 = 2\sqrt{5} \)
\( \therefore (\alpha + \beta - \alpha\beta + 2)^4 = 400 \)

Question. If \( \sin^2 A = x \) and \( \prod_{r=1}^4 \sin(rA) = ax^2+bx^3+cx^4+dx^5 \) then the value of \( a+b+c+d \) must be
Answer: \( \prod_{r=1}^4 \sin(rA) = \sin A \sin 2A \sin 3A \sin 4A \)
\( = \sin A \cdot 2\sin A\cos A \cdot (3\sin A - 4\sin^3 A) \cdot 2\sin 2A\cos 2A \)
\( = 2\sin^2 A\cos A \cdot \sin A(3 - 4\sin^2 A) \cdot 4\sin A\cos A \cdot (1 - 2\sin^2 A) \)
\( = 8x^2(1 - x)(3 - 4x)(1 - 2x) \)
\( = 24x^2 - 104x^3 + 144x^4 - 64x^5 \)
on comparing, we get a=24, b = - 104, c= 144, d= -64

Question. If \( \cot(\theta-\alpha), 3\cot\theta, \cot(\theta+\alpha) \) are in A.P and \( \theta \) is not an integral multiple of \( \frac{\pi}{2} \) then the value of \( \frac{4\sin^2\theta}{3\sin^2\alpha} \) must be
Answer: \( \because \cot(\theta-\alpha), 3\cot\theta, \cot(\theta+\alpha) \) are in AP
\( 6\cot\theta = \cot(\theta-\alpha) + \cot(\theta+\alpha) \)
\( \frac{6\cos\theta}{\sin\theta} = \frac{\sin 2\theta}{\sin(\theta+\alpha)\sin(\theta-\alpha)} \)

\( \implies \) \( 6\cos\theta \{\sin^2\theta - \sin^2\alpha\} = 2\sin^2\theta\cos\theta \)

\( \implies \) \( 3(\sin^2\theta - \sin^2\alpha) = \sin^2\theta \)
\( \therefore 2\sin^2\theta = 3\sin^2\alpha \)
or \( \frac{2\sin^2\theta}{3\sin^2\alpha} = 1 \)
\( \therefore \frac{4\sin^2\theta}{3\sin^2\alpha} = 2 \times \frac{2\sin^2\theta}{3\sin^2\alpha} = 2 \)

JEE MAINS ADVANCED

Question. If \( A+B+C=\pi \) and \( \frac{\sin 2A + \sin 2B + \sin 2C}{\sin A + \sin B + \sin C} = \lambda \sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right) \), then the value of \( \lambda \) must be
Answer: \( \because \sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C = 32\sin(A/2)\sin(B/2)\sin(C/2)\cos(A/2)\cos(B/2)\cos(C/2) \) and
\( \sin A + \sin B + \sin C = 4\cos(A/2)\cos(B/2)\cos(C/2) \)
(from conditional identities)
\( \therefore \frac{\sin 2A + \sin 2B + \sin 2C}{\sin A + \sin B + \sin C} = 8\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right) \)
on comparing, we get \( \lambda = 8 \)

Question. If \( a\tan\alpha + \sqrt{a^2-1}\tan\beta + \sqrt{a^2+1}\tan\gamma = 2a \), where a is constant and \( \alpha,\beta,\gamma \) are variable angles, by using \( |\vec{U}\cdot\vec{V}| \le |\vec{U}||\vec{V}| \). The least value of \( (\tan^2\alpha + \tan^2\beta + \tan^2\gamma) \) is
Answer: We have
\( \left( a\tan\beta - \sqrt{a^2-1}\tan\alpha \right)^2 + \left( \sqrt{a^2+1}\tan\beta - \sqrt{a^2-1}\tan\gamma \right)^2 + \left( a\tan\gamma - \sqrt{a^2+1}\tan\alpha \right)^2 \ge 0 \)

\( \implies \) \( \{a^2 + a^2 - 1 + a^2 + 1\}(\tan^2\alpha + \tan^2\beta + \tan^2\gamma) - \left\{ a\tan\alpha + \sqrt{a^2-1}\tan\beta + \sqrt{a^2+1}\tan\gamma \right\}^2 \ge 0 \)

\( \implies \) \( \tan^2\alpha + \tan^2\beta + \tan^2\gamma \ge \frac{4}{3} \)

Question. The maximum value of the expression \( \frac{1}{\sin^2\theta + 3\sin\theta\cos\theta + 5\cos^2\theta} \) is
Answer: \( \frac{1}{4\cos^2\theta + 1 + \frac{3}{2}\sin 2\theta} \)

\( \implies \) \( \frac{1}{2[1 + \cos 2\theta] + 1 + \frac{3}{2}\sin 2\theta} \)
lies between \( \frac{1}{2} \) to \( \frac{11}{2} \)
\( \therefore \) maximum value is 2.
Minimum value of \( 1 + 4\cos^2\theta + 3\sin\theta\cos\theta \)
\( 1 + \frac{4(1 + \cos 2\theta)}{2} + \frac{3}{2}\sin 2\theta \)
\( = 1 + 2 + 2\cos 2\theta + \frac{3}{2}\sin 2\theta \)
\( 3 + 2\cos 2\theta + \frac{3}{2}\sin 2\theta \)
\( \therefore = 3 - \sqrt{4 + \frac{9}{4}} = 3 - \frac{5}{2} = \frac{1}{2} \)
So maximum value of
\( \frac{1}{4\cos^2\theta + 1 + \frac{3}{2}\sin 2\theta} \) is 2

Question. The positive integer value of \( n > 3 \) satisfying the equation \( \frac{1}{\sin\left(\frac{\pi}{n}\right)} = \frac{1}{\sin\left(\frac{2\pi}{n}\right)} + \frac{1}{\sin\left(\frac{3\pi}{n}\right)} \) is
Answer: \( \frac{1}{\sin\frac{\pi}{n}} - \frac{1}{\sin\frac{3\pi}{n}} = \frac{1}{\sin\frac{2\pi}{n}} \)

\( \implies \) \( \frac{\sin\frac{3\pi}{n} - \sin\frac{\pi}{n}}{\sin\frac{\pi}{n}\sin\frac{3\pi}{n}} = \frac{1}{\sin\frac{2\pi}{n}} \)
\( = \frac{2\sin\frac{\pi}{n}\cos\frac{2\pi}{n}}{\sin\frac{\pi}{n}\sin\frac{3\pi}{n}} = \frac{\sin\frac{2\pi}{n}}{\sin\frac{2\pi}{n}} = 1 \)

\( \implies \) \( \sin\frac{4\pi}{n} = \sin\frac{3\pi}{n} \)

\( \implies \) \( \frac{4\pi}{n} + \frac{3\pi}{n} = \pi \)

\( \implies \) \( n = 7 \)

SUBJECTIVE QUESTIONS

Question. Show that \( \cos(\sin\theta) > \sin(\cos\theta), 0 \le \theta \le \frac{\pi}{2} \)
Answer: \( \sin\theta + \cos\theta \le \sqrt{2} < \frac{\pi}{2} \)

\( \implies \) \( \sin\theta < \frac{\pi}{2} - \cos\theta \)

\( \implies \) \( \cos(\sin\theta) > \cos\left(\frac{\pi}{2} - \cos\theta\right) = \sin(\cos\theta) \)

Question. If \( e^{-\frac{\pi}{2}} < \theta < e^{\frac{\pi}{2}} \) then P.T \( \cos(\ln\theta) > \ln(\cos\theta) \)?
Answer:  \( e^{-\frac{\pi}{2}} < \theta < e^{\frac{\pi}{2}} \)

\( \implies \) \( -\frac{\pi}{2} < \ln\theta < \frac{\pi}{2} \)
\( \therefore \cos(\ln\theta) > 0 \)
\( \cos\theta < 1 \)

\( \implies \) \( \ln(\cos\theta) < 0 \)
\( \therefore \cos(\ln\theta) \) is the larger

Question. Prove that \( \sin\theta + \sin 3\theta + \sin 5\theta + \dots + \sin(2n-1)\theta = \frac{\sin^2 n\theta}{\sin\theta} \)
Answer:  \( \frac{\sin\left[n\left(\frac{2\theta}{2}\right)\right]}{\sin\left(\frac{2\theta}{2}\right)}\sin\left(\frac{\theta+(2n-1)\theta}{2}\right) = \frac{\sin^2 n\theta}{\sin\theta} \)

Question. prove that \( \frac{3 - \tan^2 \frac{\pi}{7}}{1 - \tan^2 \frac{\pi}{7}} = 4\cos \frac{\pi}{7} \)
Answer:  Let \( \theta = \frac{\pi}{7} \)

\( \implies \) \( 3\theta = \pi - 4\theta \)

\( \implies \) \( \sin 3\theta = \sin 4\theta \)

\( \implies \) \( 3\sin\theta - 4\sin^3\theta = 4\sin\theta\cos\theta\cos 2\theta \)

\( \implies \) \( 3 - 4\sin^2\theta = 4\cos\theta(2\cos^2\theta - 1) \)

\( \implies \) \( 8\cos^3\theta - 4\cos\theta = 4\cos^2\theta - 1 \)

\( \implies \) \( 4\cos\theta(2\cos^2\theta - 1) = 4\cos^2\theta - 1 \)

\( \implies \) \( 4\cos\theta = \frac{4\cos^2\theta - 1}{2\cos^2\theta - 1} = \frac{3 - \tan^2\theta}{1 - \tan^2\theta} \)

Question. P.T \( \frac{1}{\sin 45^\circ\sin 46^\circ} + \frac{1}{\sin 47^\circ\sin 48^\circ} + \dots + \frac{1}{\sin 133^\circ\sin 134^\circ} = \operatorname{cosec} 1^\circ \)
Answer:  \( \frac{\sin 1^\circ}{\sin X\sin(X+1)^\circ} = \cot X^\circ - \cot(X+1)^\circ = \cot 45^\circ - \cot 46^\circ \dots \)
\( = \cot 45^\circ - (\cot 46^\circ + \cot 134^\circ) + (\cot 47^\circ + \cot 133^\circ) - \dots (\dots) - \cot 90^\circ \)
\( = 1 \)

Question. In \( \Delta ABC \), if \( \cot\theta = \cot A + \cot B + \cot C \), then prove that \( \sin(A-\theta)\sin(B-\theta)\sin(C-\theta) = \sin^3\theta \)
Answer: Here, \( \cot\theta = \cot A + \cot B + \cot C \)

\( \implies \) \( \cot\theta - \cot A = \cot B + \cot C \)

\( \implies \) \( \frac{\cos\theta}{\sin\theta} - \frac{\cos A}{\sin A} = \frac{\cos B}{\sin B} + \frac{\cos C}{\sin C} \)

\( \implies \) \( \frac{\cos\theta\sin A - \cos A\sin\theta}{\sin\theta\sin A} = \frac{\cos B\sin C + \sin B\cos C}{\sin B\sin C} \)

\( \implies \) \( \frac{\sin(A-\theta)}{\sin A\sin\theta} = \frac{\sin(B+C)}{\sin B\sin C} \)

\( \implies \) \( \sin(A-\theta) = \frac{\sin^2 A\sin\theta}{\sin B\sin C} \dots (i) \)
Similarly, \( \sin(B-\theta) = \frac{\sin^2 B\sin\theta}{\sin A\sin C} \dots (ii) \)
and \( \sin(C-\theta) = \frac{\sin^2 C\sin\theta}{\sin A\sin B} \dots (iii) \)
Multiplying (i), (ii) and (iii), we get,
\( \sin(A-\theta)\cdot\sin(B-\theta)\cdot\sin(C-\theta) = \sin^3\theta. \)

Question. In \( \Delta ABC \), \( \cot A + \cot B + \cot C = 0 \) then P.T \( \cos A \cos B \cos C = -1 \)
Answer:  \( \cot B + \cot C = -\cot A \)

\( \implies \) \( \frac{\sin(B+C)}{\sin B\sin C} = \frac{-\cos A}{\sin A} \)
\( \frac{\sin^2 A}{\sin B\sin C} = -\cos A \)

Question. Show that sum of the series \( \operatorname{cosec}\theta + \operatorname{cosec}2\theta + \operatorname{cosec}4\theta + \dots n\text{ terms} = \cot\frac{\theta}{2} - \cot 2^{n-1}\theta \)
Answer:  \( \frac{1}{\sin\theta} = \frac{\sin(\theta/2)}{\sin\theta\sin(\theta/2)} = \frac{\sin(\theta-\theta/2)}{\sin\theta\sin(\theta/2)} \)
\( = \cot(\theta/2) - \cot\theta \)

Question. P.T \( \tan^2 \frac{\pi}{7} + \tan^2 \frac{2\pi}{7} + \tan^2 \frac{3\pi}{7} = 21 \)
Answer:  Let \( \theta = \frac{n\pi}{4} (n=1,2,\dots,7) \)
\( 4\theta = n\pi - 3\theta \)
\( \tan 4\theta = -\tan 3\theta \)

\( \implies \) \( t^7 - 21t^5 + 35t^3 - 7t = 0 \)
\( t = \tan\left(\frac{n\pi}{7}\right), n=1,2,3,4,5,6,7 \)
\( t^6 - 21t^4 + 35t^2 - 7 = 0, \quad t^2 = X \, (t \neq 0) \)
\( X^3 - 21X^2 + 35X - 7 = 0 \)
Roots are \( \tan^2\frac{\pi}{7}, \tan^2\frac{2\pi}{7}, \tan^2\frac{3\pi}{7} \)
\( \tan^2\frac{5\pi}{7} = \tan^2\frac{2\pi}{7}, \tan^2\frac{4\pi}{7} = \tan^2\frac{3\pi}{7}, \)
\( \tan^2\frac{6\pi}{7} = \tan^2\frac{\pi}{7} \)
sum = 21

Question. In triangle ABC given \( \tan(A-B) + \tan(B-C) + \tan(C-A) = 0 \) then show that the triangle is isosceles triangle
Answer:  \( \frac{X-Y}{1+XY} + \frac{Y-Z}{1+YZ} + \frac{Z-X}{1+ZX} = 0 \)

\( \implies \) \( (X-Y)(Y-Z)(Z-X) = 0 \)

Question. If \( \frac{\cos\alpha}{\cos\beta} + \frac{\sin\alpha}{\sin\beta} = -1 \) then show that the value of \( \frac{\cos^3\beta}{\cos\alpha} + \frac{\sin^3\beta}{\sin\alpha} = 1 \)
Answer:  \( \cos\alpha = t\cos\beta \)

\( \implies \) \( \sin\alpha = -(1+t)\sin\beta \)

\( \implies \) \( 1 - t^2\cos^2\beta = (1+t)^2\sin^2\beta \)

\( \implies \) \( \sin^2\beta = \frac{1-t^2}{2t+1} \)

\( \implies \) \( \cos^2\beta = \frac{2t+t^2}{1+2t} \)

Question. Prove that \( \sum_{r=1}^n \tan r\alpha \tan(r+1)\alpha = \cot\alpha \tan(n+1)\alpha - n - 1 \)
Answer: \( \tan r\alpha \cdot \tan(r+1)\alpha + 1 \)
\( = \frac{\sin r\alpha\sin(r+1)\alpha + \cos r\alpha\cos(r+1)\alpha}{\cos r\alpha\cos(r+1)\alpha} \)
\( = \frac{\cos\alpha}{\cos r\alpha\cos(r+1)\alpha} \)
\( = \cot\alpha \cdot \frac{\sin\alpha}{\cos r\alpha\cos(r+1)\alpha} \)
\( = \cot\alpha \cdot \frac{\sin\{(r+1)\alpha - r\alpha\}}{\cos r\alpha\cos(r+1)\alpha} \)
\( = \cot\alpha \cdot \frac{\sin(r+1)\alpha\cos(r\alpha) - \cos(r+1)\alpha\sin(r\alpha)}{\cos r\alpha\cos(r+1)\alpha} \)
\( = \cot\alpha \cdot \{\tan(r+1)\alpha - \tan(r\alpha)\} \)
\( \therefore \sum_{r=1}^n \tan r\alpha \tan(r+1)\alpha + \sum_{r=1}^n 1 \)
\( = \cot\alpha\sum_{r=1}^n\{\tan(r+1)\alpha - \tan r\alpha\} \)
or \( \sum_{r=1}^n \tan r\alpha \tan(r+1)\alpha + n \)
\( = \cot\alpha \cdot \{\tan(n+1)\alpha - \tan\alpha\} \)
\( \therefore LHS = -n + \cot\alpha\{\tan(n+1)\alpha - \tan\alpha\} \)
\( = -n + \cot\alpha \cdot \tan(n+1)\alpha - \cot\alpha \cdot \tan\alpha \)
\( = \cot\alpha \cdot \tan(n+1)\alpha - n - 1 = RHS. \)

Question. In \( \Delta ABC \), \( 3\sin A + 4\cos B = 6 \), \( 4\sin B + 3\cos A = 1 \) then show that the number of possible values of \( \angle C \) is one
Answer: Square - ADD
\( \sin(A+B) = \frac{1}{2} \)
\( \therefore A + B = 150^\circ \)

\( \implies \) \( \angle C = 30^\circ \)
\( A + B = 30^\circ \)

\( \implies \) \( \angle C = 150^\circ \)

\( \implies \) \( A < 30^\circ \)
(or)

\( \implies \) \( 3\sin A < 3/2 \)

\( \implies \) \( 3\sin A + 4\cos B < \frac{3}{2} + 4 \cdot 1 < 6 \)
It is contradiction

Question. Let \( a^2+b^2 = \alpha^2+\beta^2 = 2 \) then show that the maximum value of \( S = (1-a)(1-b) + (1-\alpha)(1-\beta) \) is 8
Answer: \( a = \sqrt{2}\cos\theta, b = \sqrt{2}\sin\theta \)
\( \alpha = \sqrt{2}\cos\phi, \beta = \sqrt{2}\sin\phi \)

\( \implies \) \( S = 2 - 2[\sin(\theta+\pi/4) + \sin(\phi+\pi/4)] + 2[\cos(\theta-\phi)] \)
= 2 + 4 + 2 = 8

Question. If \( 0 < x < y < \frac{\pi}{2} \), show that \( x - \sin x < y - \sin y \)
Answer:
\( \sin y - \sin x = 2\sin\frac{y - x}{2}\cos\frac{y + x}{2} < 2\sin\frac{y - x}{2} \)
\( \therefore \sin y - \sin x < 2.\frac{y - x}{2} \sin ce \sin \theta < \theta \)
\( \therefore x - \sin x < y - \sin y \)

Question. Let a, b, c, d be real numbers such that a + b + c + d = 10, if the minimum value of \( a^2 \cot 9^\circ + b^2 \cot 27^\circ + c^2 \cot 63^\circ + d^2 \cot 81^\circ \) is \( \sqrt{n} \) (\( n \in N \)) then find n
Answer: If \( a_1, a_2, ......, a_n \) and \( b_1, b_2, ......, b_n \) are 2n real numbers, then
\( (a_1 b_1 + a_2 b_2 + ........ + a_n b_n)^2 \)
\( \leq \left( a_1^2 + a_2^2 + ...... + a_n^2 \right) \left( b_1^2 + b_2^2 + ...... + b_n^2 \right) \) .... (1)
Let \( a_1 = a\sqrt{\cot 9^\circ} \), \( a_2 = b\sqrt{\cot 27^\circ} \), \( a_3 = c\sqrt{\cot 63^\circ} \), \( a_4 = d\sqrt{\cot 81^\circ} \)
and \( b_1 = \sqrt{\tan 9^\circ} \), \( b_2 = \sqrt{\tan 27^\circ} \), \( b_3 = \sqrt{\tan 63^\circ} \), \( b_4 = \sqrt{\tan 81^\circ} \)
Now using (1), we get
\( (a + b + c + d)^2 \leq (a^2 \cot 9^\circ + b^2 \cot 27^\circ + c^2 \cot 63^\circ + d^2 \cot 81^\circ) \)
\( (\tan 9^\circ + \tan 27^\circ + \tan 63^\circ + \tan 81^\circ) \) ... (2)
But \( a + b + c + d = 10 \) (Given)
and \( (\tan 9^\circ + \tan 81^\circ) + (\tan 27^\circ + \tan 63^\circ) \)
\( = \frac{1}{\sin 9^\circ \cos 9^\circ} + \frac{1}{\sin 27^\circ \cos 27^\circ} = \frac{2}{\sin 18^\circ} + \frac{2}{\sin 54^\circ} \)
\( = \frac{2}{\frac{\sqrt{5} - 1}{4}} + \frac{2}{\frac{\sqrt{5} + 1}{4}} \)
\( = 8\left[ \frac{(\sqrt{5} + 1) + (\sqrt{5} - 1)}{4} \right] = 4\sqrt{5} \)
From (2), we get
\( 100 \leq 4\sqrt{5} (a^2 \cot 9^\circ + b^2 \cot 27^\circ + c^2 \cot 63^\circ + d^2 \cot 81^\circ) \)

\( \implies \) \( a^2 \cot 9^\circ + b^2 \cot 27^\circ + c^2 \cot 63^\circ + d^2 \cot 81^\circ \geq \frac{25}{\sqrt{5}} = 5\sqrt{5} = \sqrt{125} = \sqrt{n} \)
Hence n = 125

Question. If \( \frac{\cos \alpha}{\cos A} + \frac{\sin \alpha}{\sin A} = \frac{\cos \beta}{\cos A} + \frac{\sin \beta}{\sin A} = 1 \), where \( \alpha \) and \( \beta \) do not differ by an even multiple of \( \pi \), prove that \( \frac{\cos \alpha \cos \beta}{\cos^2 A} + \frac{\sin \alpha \sin \beta}{\sin^2 A} = -1 \).
Answer: \( \alpha \) and \( \beta \) are the roots of the equation
\( \frac{\cos \theta}{\cos A} + \frac{\sin \theta}{\sin A} = 1 \)
or, \( \left( \frac{\cos \theta}{\cos A} \right)^2 = \left( 1 - \frac{\sin \theta}{\sin A} \right)^2 \)
On simplification
\( \left( \frac{\sin \theta}{\sin A} \right)^2 - 2\left( \frac{\sin \theta}{\sin A} \right) \cos^2 A - \sin^2 A = 0 \)
Clearly, \( \frac{\sin \alpha}{\sin A} \) and \( \frac{\sin \beta}{\sin A} \) are the roots of this equation
\( \therefore \frac{\sin \alpha}{\sin A} \frac{\sin \beta}{\sin A} = -\sin^2 A \)

\( \implies \) \( \frac{\sin \alpha \sin \beta}{\sin^2 A} = -\sin^2 A \)
Similarly, by making a quadratic in \( \frac{\cos \theta}{\cos A} \), we obtain
\( \frac{\cos \alpha \cos \beta}{\cos A \cos A} = -\cos^2 A \)

\( \implies \) \( \frac{\cos \alpha \cos \beta}{\cos^2 A} = -\cos^2 A \)
Adding the above equations
\( \frac{\cos \alpha \cos \beta}{\cos^2 A} + \frac{\sin \alpha \sin \beta}{\sin^2 A} \)
\( = -(\cos^2 A + \sin^2 A) = -1 \)

Question. A right angle is divided into three positive parts \( \alpha, \beta \) and \( \gamma \). Prove that for all possible divisions \( \tan \alpha + \tan \beta + \tan \gamma > 1 + \tan \alpha \tan \beta \tan \gamma \)
Answer: \( y = \tan \alpha + \tan \beta + \tan \gamma - \tan \alpha \tan \beta \tan \gamma \)

\( \implies \) \( y = \frac{\sin(\alpha + \beta + \gamma)}{\cos \alpha \cos \beta \cos \gamma} \)

\( \implies \) \( y = \frac{1}{\cos \alpha \cos \beta \cos \gamma} \left[ \because \alpha + \beta + \gamma = \frac{\pi}{2} \right] \)

\( \implies \) \( y = \frac{2}{(\cos(\beta + \gamma) + \cos(\beta - \gamma)) \cos \alpha} \)

\( \implies \) \( y = \frac{2}{(\sin \alpha + \cos(\beta - \gamma)) \cos \alpha} \)
For given value of \( \alpha \) and varying \( \beta \) and \( \gamma \), we find that y is minimum if \( \cos(\beta - \gamma) \) is maximum i.e. \( \beta = \gamma \)
Similarly, for given \( \beta \), y is minimum if \( \alpha = \gamma \)
Thus, y is minimum, if \( \alpha = \beta = \gamma = \frac{\pi}{6} \)