Question

prove that (cosecx -cotx) ^2 =1-cosx/1+cosx​

Answer 1

ㅤ$\;\;\underline{\textbf{\textsf{ Given :-}}}$

• $\rm \bigg[cosecx - cotx\bigg]^{2} = \dfrac{1 - cosx}{1 + cosx}$

ㅤ$\;\;\underline{\textbf{\textsf{ To Prove :-}}}$

• L.H.S = R.H.S

ㅤ$\;\;\underline{\textbf{\textsf{ Proof :-}}}$

LHS

$\leadsto \rm \bigg[cosecx - cotx \bigg]^{2}$

$\leadsto \rm cosec^{2}x + cot^{2}x - 2\big[cosecx\big]\big[cotx\big]$

$\leadsto \rm \dfrac{1}{sin^2x} + \dfrac{cos^2x}{sin^2x} - 2\bigg[\dfrac{1}{sinx}\bigg]\bigg[\dfrac{cosx}{sinx} \bigg]$

$\leadsto \rm \dfrac{1}{sin^2x} + \dfrac{cos^2x }{sin^2x} - \dfrac{2cosx}{sin^2x}$

$\leadsto \rm \dfrac{1 + cos^2x - 2cosx}{sin^2x}$

$\leadsto \rm \dfrac{\ \big[1 - cosx\big]^2}{sin^2x}$

$\leadsto \rm \dfrac{\ \big[1 - cosx\big]^2}{1 - cos^2x}$

$\leadsto \rm \dfrac{\ \big[1 - cosx\big]^2}{\big[1 - cosx\big] \big[1 + cosx\big]}$

$\leadsto \rm \dfrac{1 - cosx}{1 + cosx}$

ㅤ$\;\;\underline{\textbf{\textsf{Hence -}}}$

$\leadsto \ {\boxed{\tt{LHS = RHS}}}$

$\therefore{ \underline{\bf{(Proved)}}}$

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ㅤ$\;\;\underline{\textbf{\textsf{Need to know -}}}$

• (a - b)² = a² + b² - 2ab

• 1/cosec²x = sin²x

• 1/cot²x = cos²x/sin²x

• cosecx = 1/sinx

• cotx = cosx/sinx

•sin²x = 1 - cos²x

•a² - b² = (a + b)(a - b)

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