Question

prove that (1+cosA+tanA)(sinA-cosA)/sec³A-cosec³A=sin²Acos²A​

Answer 1

Step-by-step explanation:

Question:-

$\sf Prove \: that :-$

$\sf \dfrac{(1 + cotA + tanA)(sinA - cosA)}{sec^3A - cosec^3A} = sin^2A \: cos^2 A$

Proof:-

$\sf LHS = \dfrac{(1 + cotA + tanA)(sinA - cosA)}{sec^3A - cosec^3A}$

$\sf = \dfrac{ \bigg(1 + \dfrac{cosA}{sinA} + \dfrac{sinA}{cosA} \bigg) \big(sinA - cosA \big)}{sec^3A - cosec^3A}$

$\sf = \dfrac{ \bigg( \dfrac{sinA \: cosA + {cos}^{2}A + {sin}^{2}A}{sinA \: cosA}</p><p> \bigg) \big(sinA - cosA \big)}{ \dfrac{1}{cos^3A} - \dfrac{1}{sin^3A}}$

$\sf = \dfrac{ \bigg( \dfrac{sinA \: cosA + {cos}^{2}A + {sin}^{2}A}{sinA \: cosA}</p><p> \bigg) \big(sinA - cosA \big)}{ \dfrac{sin^3A -cos^3A}{cos^3A \: sin^3A} }$

$\sf = \dfrac{ ( sinA \: cosA + {cos}^{2}A + {sin}^{2}A) \big(sinA - cosA \big)}{ sinA \: cosA } \times \dfrac{sin^3A \: cos^3A}{ sin^3A - cos^3A }$

$\sf = \dfrac{ {sin}^{3}A - {cos}^{3} A }{ sinA \: cosA } \times \dfrac{sin^3A \: cos^3A }{ sin^3A - cos^3A }$

$\sf = \dfrac{(sin^2A \: cos^2A)(sinA \: cosA) }{ sinA \: cosA}$

$\sf = sin^2A \: cos^2A$

LHS = RHS

Hence Proved