PUC Science Class 11 - Karnataka Board PUC Question Bank Solutions

Show that: \[3 \left( \sin x - \cos x \right)^4 + 6 \left( \sin x + \cos \right)^2 + 4 \left( \sin^6 x + \cos^6 x \right) = 13\]

Show that: \[2 \left( \sin^6 x + \cos^6 x \right) - 3 \left( \sin^4 x + \cos^4 x \right) + 1 = 0\]

Prove that: \[\cos^6 A - \sin^6 A = \cos 2A\left( 1 - \frac{1}{4} \sin^2 2A \right)\]

Prove that: \[\cos^6 A - \sin^6 A = \cos 2A\left( 1 - \frac{1}{4} \sin^2 2A \right)\]

Prove that:\[\tan\left( \frac{\pi}{4} + x \right) + \tan\left( \frac{\pi}{4} - x \right) = 2 \sec 2x\]

Prove that: \[\cot^2 x - \tan^2 x = 4 \cot 2 x  \text{ cosec }  2 x\]

Prove that: \[\cos 4x - \cos 4\alpha = 8 \left( \cos x - \cos \alpha \right) \left( \cos x + \cos \alpha \right) \left( \cos x - \sin \alpha \right) \left( \cos x + \sin \alpha \right)\]

Prove that \[\sin 3x + \sin 2x - \sin x = 4 \sin x \cos\frac{x}{2} \cos\frac{3x}{2}\]

Prove that: \[\cot \frac{\pi}{8} = \sqrt{2} + 1\]

 If \[\cos x = - \frac{3}{5}\]  and x lies in the IIIrd quadrant, find the values of \[\cos\frac{x}{2}, \sin\frac{x}{2}, \sin 2x\] .

 If  \[\cos x = - \frac{3}{5}\]  and x lies in IInd quadrant, find the values of sin 2x and \[\sin\frac{x}{2}\] .

If  \[\sin x = \frac{\sqrt{5}}{3}\] and x lies in IInd quadrant, find the values of \[\cos\frac{x}{2}, \sin\frac{x}{2} \text{ and }  \tan \frac{x}{2}\] . 

 If 0 ≤ x ≤ π and x lies in the IInd quadrant such that  \[\sin x = \frac{1}{4}\]. Find the values of \[\cos\frac{x}{2}, \sin\frac{x}{2} \text{ and }  \tan\frac{x}{2}\]

 If \[\cos x = \frac{4}{5}\]  and x is acute, find tan 2x 

 If \[\sin x = \frac{4}{5}\] and \[0 < x < \frac{\pi}{2}\]

If \[\text{ tan } x = \frac{b}{a}\] , then find the value of \[\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}}\] . 

If \[\tan A = \frac{1}{7}\]  and \[\tan B = \frac{1}{3}\] , show that cos 2A = sin 4B 

Prove that:  \[\cos 7°  \cos 14° \cos 28° \cos 56°= \frac{\sin 68°}{16 \cos 83°}\]

Prove that: \[\cos\frac{2\pi}{15} \cos\frac{4\pi}{15} \cos \frac{8\pi}{15} \cos \frac{16\pi}{15} = \frac{1}{16}\]