Question

cosA-sinA+1/cosA+sinA-1=cosecA+cotA​

Answer 1

From L.H.S,

$\sf \dfrac{cos A \: - sin A + 1}{cos + sin - 1} \\ \\ \sf \implies \: \dfrac{(cos A \: - sin A + 1)(cos A + sin A + 1)}{(cos A + sin A - 1)(cos A + sin A + 1)} \\ \\ \sf \implies \: \dfrac{(cos A \: + 1 - sin A )(cos A +1 + sin A )}{(cos A + sin A - 1)(cos A + sin A + 1)} \\ \\ \sf \implies \: \dfrac{(cos A + 1) {}^{2} - (sin A) {}^{2} }{(cos A + sin A) {}^{2} - (1) {}^{2} } \\ \\ \sf \implies \: \dfrac{cos A {}^{2} + 2cos A \: + 1 - sin A {}^{2} }{cosbA {}^{2} + sin A {}^{2} + 2 \: sin A \: cos A \: - 1 } \\ \\ \sf \implies \: \dfrac{cos A {}^{2} + 2cos A \: + cos A {}^{2} }{1 + 2sin A \: cos A \: - 1} \: \: \: ... \{1 = cos A {}^{2} + sin A {}^{2} \} \: \\ \\ \sf \implies \: \dfrac{2cos A {}^{2} + 2cos A}{2 \: sin A \: cos A} \\ \\ \sf \implies \: \bigg( \dfrac{2 \: cos A {}^{2} }{2 \: sin A \: cos A} \bigg) + \bigg( \dfrac{2 \: cos A}{2 \: sin A \: cos A} \bigg) \\ \\ \sf \implies \: \dfrac{cos A}{sin A} + \dfrac{1}{sin A} \\ \\ \sf \implies \: cot A \: + cosec A \\ \\ \sf \implies \: cosec A \: + \: cot A = R.H.S.$

Hence proved!

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$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$

Answer 2

$\huge\mathcal{{\colorbox{navy}{{\colorbox{lightblue}{Answer}}}}}$

LHS = ( cosA-sinA+1)/(cosA+sinA-1)

divide numerator and denominator with

sinA , we get

= ( cotA - 1 + cosecA)/( cotA+1 - cosecA )

= [cotA+cosecA- 1 ]/[ 1 - cosecA + cotA ]

=[(cotA+cosecA)-(cosec²A-cot²A)]/(1-cosecA+cotA)

=[(cotA+cosecA)-(cosecA+cotA)(cosecA-cotA)]/(1-cosecA+cotA)

= [(cosecA+cotA)(1-cosecA+cotA)]/(1-cosecA+cotA)

= cosecA + cotA

= RHS