Question

(1+cotA+tanA)(sinA-cosA)=secA/cosec²A-cosecA/sec²A​

Answer 1

$\huge\boxed{\underline{\mathcal{\red{A}\green{N}\pink{S}\orange{W}\blue{E}\pink{R}}}}$

${\bf{\green{LHS= (1+cotA+tanA) } (sinA-cosA)}}}$

${\bf{\red{=sinA+sinAcotA+sinAtanA-cosA-cosAcotA-cosAtanA}}}$

${\bf{\green{=sinA+sinAcosA/sinA+sinAsinA/cosA-cosA-cosAcosA/sinA-cosAsinA/cosA</p><p>}}}$

${\bf{\pink{=sinA+cosA+sin^2A/cosA-cosA-cos^2A/sinA-sinA}}}$

${\bf{\green{=sin^2A/cosA-cos^2A/sinA}}}$

${\bf{\orange{=sinA(sinA/cosA)cosA(cosA/sinA) -(1)}}}$

${\bf{\blue{=1/cosecA(1/cosecA/1/secA)-1/secA(1/secA/1/cosecA)}}}$

${\bf{\red{=(1/cosec^2A)(secA)-(1/sec^2)(cosecA)}}}$

${\bf{\green{=secA/cosec^2A-cosecA/sec^2A -(2)}}}$

from (1)

${\bf{\pink{=sinAtanA-cosAcotA -(3)}}}$

from 2 and 3

${\bf{\blue{=RHS}}}$

From 3

${\bf{\orange{=sinAsinA/cosA-cosAcosA/sinA}}}$

${\bf{\green{=secA/cosec^2A-sec^2A/cosecA}}}$

Answer 2

${\bf{\orange{FOLLOW ME}}}$

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