Answer & Solution

Answer: Option B

Solution:

If in the any question componendo and dividendo already used as
$$\frac{{a + b}}{{a - b}} = \frac{m}{n}$$
If second time you also want to apply componendo dividendo rule then result will be
$$\eqalign{ & \frac{a}{b} = \frac{{m + n}}{{m - n}} \cr & \Leftrightarrow \frac{{\sin \theta + \cos \theta }}{{{\text{sin}}\theta - \cos \theta }} = 3 \cr & \Leftrightarrow \sin \theta + {\text{cos}}\theta = 3\sin \theta - 3{\text{cos}}\theta \cr & \Leftrightarrow 2\sin \theta = 4{\text{cos}}\theta \cr & \Leftrightarrow \frac{{\sin \theta }}{{{\text{cos}}\theta }} = \frac{2}{1} \cr & \Leftrightarrow {\text{tan}}\theta = \frac{2}{1} = \frac{{\text{P}}}{{\text{B}}} \cr & \left[ {\therefore {\text{sin}}\theta = \frac{{\text{P}}}{{\text{H}}} = \frac{2}{{\sqrt 5 }}{\text{ and cos}}\theta = \frac{{\text{B}}}{{\text{H}}} = \frac{1}{{\sqrt 5 }}} \right] \cr} $$
Trigonometry mcq solution image

$$\eqalign{ & \Leftrightarrow {\sin ^4}\theta - {\text{co}}{{\text{s}}^4}\theta \cr & \Leftrightarrow \left( {{{\sin }^2}\theta + {\text{co}}{{\text{s}}^2}\theta } \right)\left( {{{\sin }^2}\theta - {\text{co}}{{\text{s}}^2}\theta } \right) \cr & \Leftrightarrow 1\left( {{{\sin }^2}\theta - {\text{co}}{{\text{s}}^2}\theta } \right) \cr & \Leftrightarrow {\left( {\frac{2}{{\sqrt 5 }}} \right)^2} - {\left( {\frac{1}{{\sqrt 5 }}} \right)^2} \cr & \Leftrightarrow \frac{4}{5} - \frac{1}{5} \cr & \Leftrightarrow \frac{3}{5} \cr} $$

+ sin - = 3, then the value of ^4 - co s^4 is?

Solution