Question

prove the following​

Answer 1

$\sf{\orange{\boxed{\bold{Solution}}}}$

⠀⠀⠀⠀

$\sf: \implies\: {\bold{ tan^2\theta - Sin^2\theta = tan^2\theta . sin^2\theta}}$

⠀⠀⠀⠀

$\sf{\red{\boxed{\bold{tan^2\theta = \dfrac{Sin^2\theta}{Cos^2\theta}}}}}$

⠀⠀⠀⠀

$\sf: \implies\: {\bold{ \dfrac{sin^2\theta}{cos^2\theta} - Sin^2\theta = tan^2\theta . sin^2\theta}}$

⠀⠀⠀⠀

$\sf: \implies\: {\bold{ \dfrac{Sin^2\theta - cos^2\theta . Sin^2\theta}{Cos^2\theta} = tan^2\theta . sin^2\theta}}$

⠀⠀⠀⠀

$\sf: \implies\: {\bold{ \dfrac{Sin^2\theta ( 1 - cos^2\theta )}{Cos^2\theta} = tan^2\theta . sin^2\theta}}$

⠀⠀⠀⠀

$\sf{\red{\boxed{\bold{1 - cos^2\theta = Sin^2\theta }}}}$

⠀⠀⠀⠀

$\sf: \implies\: {\bold{ \dfrac{Cos^2\theta.Sin^2\theta}{Cos^2\theta} = tan^2\theta . sin^2\theta}}$

⠀⠀⠀⠀

$\sf{\red{\boxed{\bold{tan^2\theta = \dfrac{sin^2\theta}{cos^2\theta}}}}}$

⠀⠀⠀⠀

$\sf: \implies\: {\bold{Tan^2\theta . Sin^2\theta= tan^2\theta . sin^2\theta}}$

⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀....Hence Proved

Answer 2

$\sf{\boxed{\boxed{\large{\red{\mathsf{ANSWER}}}}}}$

⠀⠀⠀⠀

$\sf{\bold{ tan^2\theta - Sin^2\theta = tan^2\theta.sin^2\theta}}$

⠀⠀⠀⠀

$\sf{\boxed{\boxed{\large{\red{\mathsf{LHS}}}}}}$

⠀⠀⠀⠀

$\sf: \longrightarrow{\bold{tan^2\theta - Sin^2\theta}}$⠀⠀⠀⠀

⠀⠀⠀⠀

• $\sf{{\boxed{\bold{tan^2\theta = \dfrac{Sin^2\theta}{Cos^2\theta}}}}}$

⠀⠀⠀⠀

$\sf: \longrightarrow\: {\bold{ \dfrac{sin^2\theta}{cos^2\theta} - Sin^2\theta }}$

⠀⠀⠀⠀

$\sf: \longrightarrow\: {\bold{ \dfrac{Sin^2\theta - cos^2\theta . Sin^2\theta}{Cos^2\theta} }}$

⠀⠀⠀⠀

$\sf: \longrightarrow\: {\bold{ \dfrac{Sin^2\theta ( 1 - cos^2\theta )}{Cos^2\theta} }}$

⠀⠀⠀⠀

• $\sf{{\boxed{\bold{1 - cos^2\theta = Sin^2\theta }}}}$

⠀⠀⠀⠀

$\sf: \longrightarrow\: {\bold{ \dfrac{Cos^2\theta.Sin^2\theta}{Cos^2\theta}} }$

⠀⠀⠀⠀

• $\sf{{\boxed{\bold{tan^2\theta = \dfrac{sin^2\theta}{cos^2\theta}}}}}$

$\sf: \longrightarrow\: {\bold{Tan^2\theta . Sin^2\theta}}$

⠀⠀⠀⠀

$\sf{\boxed{\boxed{\large{\red{\mathsf{RHS}}}}}}$

⠀⠀⠀⠀

$\sf: \longrightarrow\: {\bold{Tan^2\theta . Sin^2\theta}}$

⠀⠀⠀⠀

$\sf{\boxed{\boxed{\large{\red{\mathsf{COMPARE}}}}}}$

⠀⠀⠀⠀

$\sf {\bold{Tan^2\theta . Sin^2\theta}}$ = $\sf {\bold{Tan^2\theta . Sin^2\theta}}$