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Question Number 10743 by okhema last updated on 24/Feb/17

s h o w t h a t \sec^{2} θ = \frac{cosec θ}{cosec θ - \sin}

Commented by ridwan balatif last updated on 24/Feb/17

\frac{cosec θ}{cosec θ - \sin θ} = \frac{\frac{1}{\sin θ}}{\frac{1}{\sin θ} - \sin θ}

= \frac{\frac{1}{\sin θ}}{\frac{1 - \sin^{2} θ}{\sin θ}}

= \frac{\frac{\frac{1}{\sin θ}}{\cos^{2} θ}}{\sin θ}

= \frac{1}{\sin θ} \times \frac{\sin θ}{\cos^{2} θ}

= \frac{1}{\cos^{2} θ}

\frac{cosec θ}{cosec θ - \sin θ} = \sec^{2} θ

remember!

\sec θ = \frac{1}{\cos θ}

cosec θ = \frac{1}{\sin θ}

Commented by okhema last updated on 24/Feb/17

y e s t h a t s w h a t i g o t b u t w a s n t s u r e i f i t s c o r r e c t

c a u s e i o n l y k n o w {s e c}^{2} θ = 1 + {t a n}^{2} θ b u t t h a n k s a n y w a y s

Answered by ridwan balatif last updated on 24/Feb/17

\sec^{2} θ = \frac{1}{\cos^{2} θ}

= (\frac{1}{1 - \sin^{2} θ}) (\frac{cosec θ}{cosec θ})

= \frac{cosec θ}{cosec θ - cosec θ \times \sin θ \times \sin θ}

= \frac{cosec θ}{cosec θ - \frac{1}{\sin θ} \times \sin θ \times \sin θ}

\sec^{2} θ = \frac{cosec θ}{cosec θ - \sin θ}

PROVED!!!!

Commented by okhema last updated on 24/Feb/17

w h y c o u l d n t y o u s t a r t f r o m t h e r i g h t \dots c a u s e t h a t s w h a t i d o b y d o i n g t h e r i g h t h a n d s i d e b u t i e n d u p w i t h \frac{1}{{c o s}^{2} θ}

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