Question

$\frac{(1-{\tan }^2\mathrm{\theta })}{(1+{\tan }^2\mathrm{\theta })}=?$
(a) (sin²θ − cos²θ)
(b) (cos²θ − sin²θ)
(c) (cot²θ − tan²θ)
(d) (tan²θ − cot²θ)

Answer 1

(b) (cos²θ − sin²θ)

$$\begin{gathered} \frac{1-{\text{tan}}^2\theta }{1+{\text{tan}}^2\theta } \\ \text{=}\frac{(1-\frac{{\text{Sin}}^2\theta }{{\text{Cos}}^2\theta })}{(1+\frac{{\text{Sin}}^2\theta }{{\text{Cos}}^2\vartheta })} \\ =\frac{\left(\frac{{\text{Cos}}^2\theta -{\text{Sin}}^2\theta }{{\text{Cos}}^2\theta }\right)}{\left(\frac{{\text{Cos}}^2\theta +{\text{Sin}}^2\theta }{{\text{Cos}}^2\theta }\right)} \\ \text{=}\frac{{\text{cos}}^2\theta -s{\text{in}}^2\theta }{{\text{cos}}^2\theta +s{\text{in}}^2\theta } \\ {\text{=cos}}^2\theta -s{\text{in}}^2\theta [\because c{\text{os}}^2\theta +s{\text{in}}^2\theta =1] \end{gathered}$$