Question

$\frac{? \\cot(90 - theta) tan \: theta }{sec {?}^{2} } - \sin( {?}^{2} theta = \frac{ \tan( {?}^{2} ) }{ \sec( {2}^ ) }$
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Answer 1

$\huge\mathcal{Answer:}$

In the given question, take LHS

$\frac{cot (90 - θ) \times tan θ}{ {sec}^{2}  θ} - {sin}^{2}  θ$

We know that,

$\boxed{cot(90 -  θ) = tan θ}$

$= \frac{tan θ \times tan θ}{ {sec}^{2} θ } - {sin}^{2}  θ$

$= \frac{ {tan}^{2}  θ}{ {sec}^{2} θ } - {sin}^{2}  θ$

$\boxed { {tan}^{2}  θ = \frac{ {sin}^{2} θ }{ {cos}^{2}  θ} \: ; {sec}^{2}  θ = \frac{ 1 }{ {cos}^{2}  θ} }$

$= (\frac{ {sin}^{2}θ }{ {cos}^{2}θ} \times {cos}^{2} θ) - {sin}^{2} θ$

${sin}^{2} θ - {sin}^{2} θ = 0$

LHS = RHS, hence proved.

$\frac{cot (90 - θ) \times tan θ}{ {sec}^{2}  θ} - {sin}^{2}  θ = 0$

Have a good evening!