Question

(1 + tan square o) cot o/cosec square o
= tan 0​

Answer 1

Step-by-step explanation:

1+ tan^2∅ = sec^2∅

=> LHS = (sec^2∅×cot∅)/cosec^2∅

= (sec^∅ × cosec∅)/cosec^2∅

= sec^∅/cosec∅ = tan∅ = RHS

hence proved .

_____________

Answer 2

$\Huge{\underline{\underline{\mathfrak{Question \colon}}}}$

$\large{ \sf{ \frac{(1 + tan {}^{2}x)cot \: x }{cosec {}^{2}x} = tan \: x}}$

$\Huge{\underline{\underline{\mathfrak{Answer \colon}}}}$

LHS

$\sf{ \frac{(1 + {tan}^{2}x)cot \: x }{ {cosec}^{2} x} }$

From sec²∅ - tan²∅ = 1 » 1 + tan²∅ = sec²∅

$\longrightarrow \: \sf{ \frac{ {sec}^{2}x .cot \: x }{cosec {}^{2} x} }$

Expressing them in their sine and cosine forms,

$\longrightarrow \: \sf{ \frac{ \frac{1}{ {cos}^{2}x }. \frac{cos \: x}{sin \: x} }{ \frac{1}{ {sin}^{2}x } } } \\ \\ \longrightarrow \: \sf{ \frac{sin {}^{2}x.cos \: x }{cos {}^{2}x.sin \: x } } \\ \\ \longrightarrow \: \sf{ \frac{sin \: x}{cos \: x} } \\ \ \ \\ \huge{\longrightarrow \: \sf{tan \: x}}$

Hence,Proved