Question

Prove the following identity. Kindly, don't spam. Irrelevant answer will be reported.​

Answer 1

Solution :



$\dfrac{1+sin}{cosecA-cotA} \: - \: \dfrac{1+sin}{cosecA+cotA} \:=\: 2(1+cot A)$

TAKE LHS

$\implies \dfrac{1+sinA}{\dfrac{1}{sinA}-\dfrac{cosA}{sinA}} \: - \:\dfrac{1-sinA}{\dfrac{1}{sinA}+\dfrac{cosA}{sinA}}$

$\implies \dfrac{1+sinA}{\dfrac{1-cosA}{sinA}}\: - \:\dfrac{1-sinA}{\dfrac{1+cosA}{sinA}}$

$\implies\dfrac{sinA(1+sinA)}{1-cosA}\:-\:\dfrac{sinA(1-sinA)}{1+cosA}$

$\implies sinA\Bigg[\dfrac{1+sinA}{1-cosA}\:-\:\dfrac{1-sinA}{1+cosA}\Bigg]$

$\implies sinA\Bigg[\dfrac{(1+sinA)(1+cosA)}{1-cos^{2}A}\:-\:\dfrac{(1-sinA)(1-cosA)}{1-cos^{2}A}\Bigg]$

$\implies sinA\Bigg[\dfrac{(1+sinA)(1+cosA)\: - \: (1-sinA)(1-cosA)}{1-cos^{2}A}\Bigg]$

$\implies sinA\Bigg[\dfrac{(1+sinA)(1+cosA)\: - \: (1-sinA)(1-cosA)}{sin^{2}A}\Bigg]$

$\implies \dfrac{(1+sinA)(1+cosA)\: - \: (1-sinA)(1-cosA)}{sinA}$

$\implies \dfrac{1+cosA+sinA+sinAcosA-(1-cosA-sinA+sinAcosA)}{sinA}$

$\implies \dfrac{1+cosA+sinA+sinAcosA-1+cosA+sinA-sinAcosA}{sinA}$

$\implies \dfrac{2cosA\:+\:2sinA}{sinA}$

$\implies 2\Bigg[\dfrac{cosA\:+\:sinA}{sinA}\Bigg]$

$\implies 2\Bigg[\dfrac{cosA}{sinA} \:+\:{sinA}{sinA}\Bigg]$

RHS

$\implies 2[CotA \:+\:1]$

LHS = RHS

Hence Proved

Answer 2

Step-by-step explanation:

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