Question

If at $\theta ={\left(11\frac{1}{4}\right)}^{0}and\frac{\cos 3\theta }{\cos \theta }+\frac{\sin 3\theta }{\sin \theta }=a\left(\sqrt{b+1}\right),$ ( where a & b are natural numbers ) then

• A. $a=2\sqrt{2}$
• B. $b=2$
• C. $a=b$
• D. $a>b$

Answer 1

The correct option is A $a=2\sqrt{2}$

Given

$\theta =11{\frac{1}{4}}^{\circ }⟹2\theta =22{\frac{1}{2}}^{\circ }$

Consider

$\frac{\cos 3\theta }{\cos \theta }$

$\frac{4{\cos }^3\theta -3\cos \theta }{\cos \theta }$

$⟹4{\cos }^2\theta -3\cdots \left(1\right)$

Consider

$\frac{\sin 3\theta }{\sin \theta }$

$\frac{3\sin \theta -4{\sin }^3\theta }{\sin \theta }$

$⟹3-4{\sin }^2\theta \cdots \left(2\right)$

From $\left(1\right)+\left(2\right)$

$⟹\frac{\cos 3\theta }{\cos \theta }+\frac{\sin 3\theta }{\sin \theta }$

$⟹4{\cos }^2\theta -3+3-4\sin 62\theta$

$⟹4({\cos }^2\theta -{\sin }^2\theta )$

$⟹4\cos 2\theta$

$({\cos }^2\theta -{\sin }^2\theta =\cos 2\theta )$

$⟹4\cos 22{\frac{1}{2}}^{\circ }$

$⟹4\sqrt{\frac{1+\cos {45}^{\circ }}{2}}$

$(\cos \theta =\sqrt{\frac{1+\cos 2\theta }{2}})$

$⟹4\sqrt{\frac{\sqrt{2}+1}{2\sqrt{2}}}$

Multiplying and diving by $4\sqrt{2}$ inside the square root

$⟹4\sqrt{\frac{(\sqrt{2}+1)4\sqrt{2}}{16}}$

$⟹\sqrt{8+4\sqrt{2}}$

$⟹2\sqrt{\sqrt{2}+2}$

$⟹2\sqrt{2}\sqrt{\frac{1}{\sqrt{2}}+1}$

Comparing with $a\left(\sqrt{b+1}\right)$

$⟹a=2\sqrt{2}$