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NCERT Exercise 3.3 Question 6 \(\cos \left ( \frac{\pi}{4} – x \right )\cos \left ( \frac{\pi}{4} – y \right ) – \sin \left ( \frac{\pi}{4} – x \right )\sin \left ( \frac{\pi}{4} – y \right ) = \sin (x+y)\)

Question 7 \(\frac{\tan \left ( \frac{\pi}{4} + x \right )}{\tan \left ( \frac{\pi}{4} – x \right )} = \left ( \frac{1 + \tan x}{1 – \tan x} \right ) ^2\)

Question 8 \(\frac{cos(\pi+x) \cos(-x)}{\sin (\pi – x) \cos \left ( \frac{\pi}{2} + x \right )} = \cot^2 x\)

Question 9 \( \cos \left ( \frac{3\pi}{2} + x \right ) cos (2 \pi + x) \left [ \cot \left ( \frac{3\pi}{2} – x \right ) + \cot (2 \pi + x) \right ] = 1\)

Question 10 \( \sin (n + 1)x \sin (n + 2)x + \cos (n + 1)x \cos (n + 2)x = \cos x\)

Derivation for

\( \sin (A+B)+\sin (A-B) = 2 \sin A \cos B \)

\( \sin (A+B)-\sin (A-B) = 2 \cos A \sin B \)

\( \cos (A+B)+\cos (A-B) = 2 \cos A \cos B \)

\( \cos (A+B)-\cos (A-B) = – 2 \sin A \sin B \)

Question 11 \( \cos \left ( \frac{3\pi}{4} + x \right ) – \cos \left ( \frac{3\pi}{4} – x \right ) = – \sqrt{2} \sin x\)

Derivation for twice angles

\(\sin 2x = 2 \sin x \cos x = \frac{2 \tan x}{1 + \tan^2 x}\)

\( \cos 2x = \cos^2 x – \sin^2 x = 1 – 2 \sin^2 x = 2 \cos^2 x – 1 = \frac{1-\tan^2 x}{1 + \tan^2 x}\)

Question 24 \( \cos 4x = 1 – 8 \sin^2 x \cos^2 x\)

Question 23 \( \tan 4x = \frac{4 \tan x (1 – \tan^2 x)}{1-6\tan^2 x + \tan^4 x}\)