Question

If $$\frac{{\sin \left( {x + y} \right)}}{{\sin \left( {x - y} \right)}} = \frac{{a + b}}{{a - b}},$$     then what is $$\frac{{\tan x}}{{\tan y}}$$  equal to ?

A. $$\frac{b}{a}$$
B. $$\frac{a}{b}$$  
C. $$ab$$
D. $$1$$
Answer :   $$\frac{a}{b}$$
Solution :
$$\frac{{\sin \left( {x + y} \right)}}{{\sin \left( {x - y} \right)}} = \frac{{a + b}}{{a - b}}$$
Applying componendo and dividendo, we get
$$\eqalign{ & \frac{{\sin \left( {x + y} \right) + \sin \left( {x - y} \right)}}{{\sin \left( {x + y} \right) - \sin \left( {x - y} \right)}} = \frac{{\left( {a + b} \right) + \left( {a - b} \right)}}{{\left( {a + b} \right) - \left( {a - b} \right)}} \cr & \Rightarrow \frac{{2\sin x \cdot \cos y}}{{2\cos x \cdot \cos y}} = \frac{{2a}}{{2b}} \cr & \Rightarrow \tan x \cdot \cot y = \frac{a}{b} \cr & \therefore \frac{{\tan x}}{{\tan y}} = \frac{a}{b} \cr} $$

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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