Question

Prove that:
tan Ѳ-cotѲ/sinѲ.cosѲ
= sec2 Ѳ – cosec2 Ѳ

Answer 1

Answer:

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Answer 2

Given : tan Ѳ-cotѲ / sinѲ.cosѲ = sec² Ѳ – cosec² Ѳ

Exigency to Prove : tan Ѳ-cotѲ / sinѲ.cosѲ = sec² Ѳ – cosec² Ѳ

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$\qquad \dag\:\:\bigg\lgroup \sf{ \dfrac{tan \theta - cot \theta }{ sin \theta \times cosec \theta } = sec^2 \theta - cos^2 \theta }\bigg\rgroup$

⠀⠀Here ,

• $\bf L.H.S \:: \sf \dfrac{tan \theta - cot \theta }{ sin \theta \times cos \theta }$
• $\bf R.H.S \:: \sf sec^2 \theta - cosec^2 \theta$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Solving \: the \: L.H.S \: \::}}$

$\bf L.H.S \:: \sf \dfrac{tan \theta - cot \theta }{ sin \theta \times cos \theta }$

$\qquad \longmapsto \: \sf \dfrac{tan \theta - cot \theta }{ sin \theta \times cos \theta }$

$\qquad \longmapsto \: \sf \dfrac{ \dfrac{ sin \theta }{cos \theta } - cot \theta }{ sin \theta \times cos \theta } \qquad \:\bigg\lgroup \sf{ tan \theta = \dfrac{ sin \theta }{cos \theta } }\bigg\rgroup$

$\qquad \longmapsto \: \sf \dfrac{ \dfrac{ sin \theta }{cos \theta } - \dfrac{ cos \theta }{sin \theta } }{ sin \theta \times cos \theta } \qquad \:\bigg\lgroup \sf{ cot \theta = \dfrac{ cos \theta }{sin \theta } }\bigg\rgroup$

$\qquad \longmapsto \: \sf \dfrac{ \dfrac{ sin \theta }{cos \theta } - \dfrac{ cos \theta }{sin \theta } }{ sin \theta \times cos \theta }$

$\qquad \longmapsto \: \sf \dfrac{ sin^2 \theta - cos ^2 \theta }{ (sin \theta \times cos \theta)^2 }$

$\qquad \longmapsto \: \sf \dfrac{ sin^2 \theta - cos ^2 \theta }{ sin^2 \theta \times cos^2 \theta }$

$\qquad \longmapsto \: \sf \dfrac{ sin^2 \theta }{ sin^2 \theta \times cos^2 \theta } - \dfrac{ cos^2 \theta }{ sin^2 \theta \times cos^2 \theta }$

$\qquad \longmapsto \: \sf \dfrac{ \cancel {sin^2 \theta} }{ \cancel {sin^2 \theta} \times cos^2 \theta } - \dfrac{\cancel {cos^2 \theta }}{ sin^2 \theta \times \cancel {cos^2 \theta} }$

$\qquad \longmapsto \: \sf \dfrac{ 1 }{ cos^2 \theta } - \dfrac{ 1 }{ sin^2 \theta}$

$\qquad \longmapsto \: \sf sec^2 \theta - \dfrac{ 1 }{ sin^2 \theta} \qquad \bigg\lgroup \sf{ sec \theta = \dfrac{ 1 \theta }{cos \theta } }\bigg\rgroup$

$\qquad \longmapsto \bf L.H.S \: \sf sec^2 \theta - cosec^2 \theta$

Therefore,

• $\bf L.H.S \:: \sf sec^2 \theta - cosec^2 \theta$
• $\bf R.H.S \:: \sf sec^2 \theta - cosec^2 \theta$

⠀⠀⠀⠀⠀$\therefore {\underline {\bf{ Hence, \:Proved \:}}}$

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