Question

Solve :
$({\sin }^4\theta -{\cos }^4\theta +1)(1+\frac{1}{{\tan }^4\theta })$=0

• A. $1$
• B. $2$
• C. $3$
• D. No solution

Answer 1

The correct option is D No solution

$\begin{array}{c}\left({\sin }^4\theta -{\cos }^4\theta +1\right)\left(1+\frac{1}{{\tan }^4\theta }\right)\\ \Rightarrow \left[{\left({\sin }^2\theta \right)}^2-{\left({\cos }^2\theta \right)}^2+1\right]\times \left(\frac{{\sin }^4\theta +{\cos }^4\theta }{{\sin }^4\theta }\right)\\ \Rightarrow \left\{\left({\sin }^2\theta +{\cos }^2\theta \right)\left({\sin }^2\theta -{\cos }^2\theta \right)+1\right\}\left\{\frac{{\left({\sin }^2\theta +{\cos }^2\theta \right)}^2-2{\sin }^2\theta \cdot {\cos }^2\theta }{{\sin }^4\theta }\right\}\\ \Rightarrow \left\{{\sin }^2\theta +{\cos }^2\theta +{\sin }^2\theta +{\cos }^2\theta \right\}\left\{\frac{1-2{\sin }^2\theta \cdot {\cos }^2\theta }{{\sin }^4\theta }\right\}\\ \Rightarrow 2{\sin }^2\theta \times \frac{\left(1-2{\sin }^2\theta \cdot {\cos }^2\theta \right)}{{\sin }^4\theta }\\ 2\left(\cos e{c}^2\theta -2{\cos }^2\theta \right)=0\end{array}$

$si{n}^{2}2\theta =2$

which is not possible