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1

WB JEE 2009

Subjective

Show that $${{\sin \theta } \over {\cos 3\theta }} + {{\sin 3\theta } \over {\cos 9\theta }} + {{\sin 9\theta } \over {\cos 27\theta }} = {1 \over 2}(\tan 27\theta - \tan \theta )$$

Answer

.

Explanation

$${{\sin \theta } \over {\cos 3\theta }} + {{\sin 3\theta } \over {\cos 9\theta }} + {{\sin 9\theta } \over {\cos 27\theta }} = {1 \over 2}(\tan 27\theta - \tan \theta )$$

Here, $${{\sin \theta } \over {\cos 3\theta }} = {1 \over 2}{{2\sin \theta \,.\,\cos \theta } \over {\cos 3\theta \,.\,\cos \theta }} = {1 \over 2}{{\sin 2\theta } \over {\cos 3\theta \,.\,\cos \theta }}$$

$$ = {1 \over 2}{{\sin (3\theta - \theta )} \over {\cos 3\theta \,.\,\cos \theta }} = {1 \over 2}(\tan 3\theta - \tan \theta )$$ ..... (i)

Similarly, $${{\sin 3\theta } \over {\cos 9\theta }} = {1 \over 2}(\tan 9\theta - \tan 3\theta )$$ (Replace $$\theta$$ by 3$$\theta$$ in (i))

$${{\sin 9\theta } \over {\cos 27\theta }} = {1 \over 2}(\tan 27\theta - \tan 9\theta )$$ (Replace $$\theta$$ by 9$$\theta$$ in (i))

$$\therefore$$ L.H.S. $$ = {1 \over 2}[(\tan 3\theta - \tan \theta ) + (\tan 9\theta - \tan 3\theta ) + (\tan 27\theta - \tan 9\theta )]$$

$$ = {1 \over 2}(\tan 27\theta - \tan \theta ) = $$ R.H.S.

2

WB JEE 2008

Subjective

Find the general solution of $$\sec \theta + 1 = (2 + \sqrt 3 )\tan \theta $$.

Answer

.

Explanation

$$\sec \theta + 1 = (2 + \sqrt 3 )\tan \theta $$

Squaring both sides, we get

$${(\sec \theta + 1)^2} = {(2 + \sqrt 3 )^2}{\tan ^2}\theta $$

$$ \Rightarrow {(\sec \theta + 1)^2} = (4 + 4\sqrt 3 + 3)({\sec ^2}\theta - 1)$$

$$ \Rightarrow {(\sec \theta + 1)^2} - (7 + 4\sqrt 3 )({\sec ^2}\theta - 1) = 0$$

$$ \Rightarrow {(\sec \theta + 1)^2} - (7 + 4\sqrt 3 )(\sec \theta + 1)(\sec \theta - 1) = 0$$

$$ \Rightarrow (\sec \theta + 1)[\sec \theta + 1 - (7 + 4\sqrt 3 )(\sec \theta - 1)] = 0$$

$$ \Rightarrow (\sec \theta + 1)[(8 + 4\sqrt 3 ) - (6 + 4\sqrt 3 )\sec \theta ] = 0$$

$$ \Rightarrow \sec \theta + 1 = 0$$ or $$(8 + 4\sqrt 3 ) - (6 + 4\sqrt 3 )\sec \theta = 0$$

If $$\sec \theta + 1 = 0$$

$$ \Rightarrow \cos \theta = - 1 = \cos \pi \Rightarrow \theta = 2n\pi \pm \pi $$ ..... (i)

If $$(8 + 4\sqrt 3 ) - (6 + 4\sqrt 3 )\sec \theta = 0$$

$$ \Rightarrow \sec \theta = {{8 + 4\sqrt 3 } \over {6 + 4\sqrt 3 }} = {{4(2 + \sqrt 3 )} \over {2\sqrt 3 (2 + \sqrt 3 )}} = {2 \over {\sqrt 3 }}$$

$$ \Rightarrow \cos \theta = {{\sqrt 3 } \over 2} = \cos {\pi \over 6}$$

$$ \Rightarrow \theta = 2n\pi \pm {\pi \over 6}$$

$$\because$$ angles obtained from $$\theta = 2n\pi - {\pi \over 6}$$ is not satisfying the given equation

$$\therefore$$ $$\theta = 2n\pi \pm \pi $$, $$2n\pi + {\pi \over 6}$$, $$n \in I$$

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