Question

If $$\cos 2x + 2\cos x = 1$$    then $${\sin ^2}x\left( {2 - {{\cos }^2}x} \right)$$    is equal to

A. $$1$$  
B. $$ - 1$$
C. $$ - \sqrt 5 $$
D. $$ \sqrt 5 $$
Answer :   $$1$$
Solution :
$$\eqalign{ & {\text{Here, }}{\cos ^2}x + \cos x - 1 = 0\,\,\,{\text{or, }}\cos x = \frac{{ - 1 + \sqrt 5 }}{2}\left\{ {\because \,\,\cos x \ne \frac{{ - 1 - \sqrt 5 }}{2} < - 1} \right\} \cr & \therefore \,\,{\cos ^2}x = {\left( {\frac{{\sqrt 5 - 1}}{2}} \right)^2} = \frac{{6 - 2\sqrt 5 }}{4} = \frac{{3 - \sqrt 5 }}{2} \cr & \therefore \,\,{\text{value}} = \left( {1 - \frac{{ 3 - \sqrt 5}}{2}} \right)\left( {2 - \frac{{3 - \sqrt 5 }}{2}} \right) = \frac{{\sqrt 5 - 1}}{2} \cdot \frac{{\sqrt 5 + 1}}{2} = 1. \cr} $$

Releted MCQ Question on
Trigonometry >> Trigonometric Ratio and Identities

Releted Question 1

If $$\tan \theta = - \frac{4}{3},$$   then $$\sin \theta $$  is

A. $$ - \frac{4}{5}{\text{ but not }}\frac{4}{5}$$
B. $$ - \frac{4}{5}{\text{ or }}\frac{4}{5}$$
C. $$ \frac{4}{5}{\text{ but not }} - \frac{4}{5}$$
D. None of these
Releted Question 2

If $$\alpha + \beta + \gamma = 2\pi ,$$    then

A. $$\tan \frac{\alpha }{2} + \tan \frac{ \beta }{2} + \tan \frac{\gamma }{2} = \tan \frac{\alpha }{2}\tan \frac{\beta }{2}\tan \frac{\gamma }{2}$$
B. $$\tan \frac{\alpha }{2}\tan \frac{\beta }{2} + \tan \frac{\beta }{2}\tan \frac{\gamma }{2} + \tan \frac{\gamma }{2}\tan \frac{\alpha }{2} = 1$$
C. $$\tan \frac{\alpha }{2} + \tan \frac{ \beta }{2} + \tan \frac{\gamma }{2} = - \tan \frac{\alpha }{2}\tan \frac{\beta }{2}\tan \frac{\gamma }{2}$$
D. None of these
Releted Question 3

Given $$A = {\sin ^2}\theta + {\cos ^4}\theta $$    then for all real values of $$\theta $$

A. $$1 \leqslant A \leqslant 2$$
B. $$\frac{3}{4} \leqslant A \leqslant 1$$
C. $$\frac{13}{16} \leqslant A \leqslant 1$$
D. $$\frac{3}{4} \leqslant A \leqslant \frac{{13}}{{16}}$$
Releted Question 4

The value of the expression $$\sqrt 3 \,{\text{cosec}}\,{\text{2}}{{\text{0}}^ \circ } - \sec {20^ \circ }$$     is equal to

A. 2
B. $$\frac{{2\sin {{20}^ \circ }}}{{\sin {{40}^ \circ }}}$$
C. 4
D. $$\frac{{4\sin {{20}^ \circ }}}{{\sin {{40}^ \circ }}}$$

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