Question

On simplifying $$\frac{{{{\sin }^3}A + \sin 3A}}{{\sin A}} + \frac{{{{\cos }^3}A - \cos 3A}}{{\cos A}},$$        we get

A. $${\sin 3A}$$
B. $${\cos 3A}$$
C. $${\sin A} + {\cos A}$$
D. $$3$$  
Answer :   $$3$$
Solution :
$$\eqalign{ & \frac{{{{\sin }^3}A + \sin 3A}}{{\sin A}} + \frac{{{{\cos }^3}A - \cos 3A}}{{\cos A}} \cr & \Rightarrow \frac{{{{\sin }^3}A + 3\,\sin A - 4\,{{\sin }^3}A}}{{\sin A}} + \frac{{{{\cos }^3}A - \left[ {4\,{{\cos }^3}A - 3\cos A} \right]}}{{\cos A}} \cr & \Rightarrow \frac{{3\sin A - 3\,{{\sin }^3}A}}{{\sin A}} + \frac{{\left( { - 3\,{{\cos }^3}A + 3\cos A} \right)}}{{\cos A}} \cr & = 3 - 3\,{\sin ^2}A - 3\,{\cos ^2}A + 3 \cr & = 6 - 3\left( {{{\cos }^2}A + {{\sin }^2}A} \right) \cr & = 6 - 3\left( 1 \right) \cr & = 3 \cr} $$

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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