Question

Question 13
If $sin\theta -cos\theta =0$, then the value of $(si{n}^4\theta +co{s}^4\theta )$ is

(A) $1$
(B) $\frac{3}{4}$
(C) $\frac{1}{2}$
(D) $\frac{1}{4}$

Answer 1

The answer is (C).
Thinking Process
Firstly, from $sin\theta -cos\theta =0$ get the value of $\theta$ after that put the value of $\theta$ in the given expression to get the desired result.

Given, $sin\theta -cos\theta =0$

$\Rightarrow sin\theta =cos\theta \Rightarrow \frac{sin\theta }{cos\theta }=1$

$\Rightarrow tan\theta =1[\because tan\theta =\frac{sin\theta }{cos\theta }andtan{45}^{\circ }=1]$

$\Rightarrow tan\theta =tan{45}^{\circ }$

$\therefore \theta ={45}^{\circ }$

Now, $si{n}^4\theta +co{s}^4\theta =si{n}^4{45}^{\circ }+co{s}^4{45}^{\circ }$

$={\left(\frac{1}{\sqrt{2}}\right)}^4+{\left(\frac{1}{\sqrt{2}}\right)}^4[\because sin{45}^{\circ }=cos{45}^{\circ }=\frac{1}{\sqrt{2}}]$

$=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$