Question

If $$x = \alpha ,\beta $$   satisfies both the equations $${\cos ^2}x + a\cos x + b = 0$$     and $${\sin ^2}x + p\sin x + q = 0$$     then the relation between $$a, b, p$$  and $$q$$ is

A. $$1 + b + {a^2} = {p^2} - q - 1$$
B. $${a^2} + {b^2} = {p^2} + {q^2}$$
C. $$2\left( {b + q} \right) = {a^2} + {p^2} - 2$$  
D. None of these
Answer :   $$2\left( {b + q} \right) = {a^2} + {p^2} - 2$$
Solution :
$$\eqalign{ & \cos \alpha + \cos \beta = - a,\cos \alpha \cdot \cos \beta = b\,\,{\text{and }}\sin \alpha + \sin \beta = - p,\sin \alpha \cdot \sin \beta = q. \cr & \therefore \,\,{a^2} + {p^2} = 2 + 2\cos \left( {\alpha - \beta } \right)\,\,{\text{and }}b + q = \cos \left( {\alpha - \beta } \right). \cr & \therefore \,\,{a^2} + {p^2} = 2 + 2\left( {b + q} \right). \cr} $$

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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