Question

Tama/1+seca-tana/1-seca=2cosecaprove
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Answer 1

solving here step wise,

tanA (1- secA) – tanA (1+secA) / 1- sec^2A

= -2tanAsecA / 1- (1 + tan^2A)

= -2tanAsecA / -tan^2A

= 2secA / tanA

= secA * cotA

= 2/ cosA * cosA / sinA

= 2/ sinA

= 2cosecA

Hence proved.

Answer 2

LHS=1+secAtanA−1−secAtanA

=tanA\big(\frac{1}{1+secA}-\frac{1}{1-secA}\big)=tanA(1+secA1−1−secA1)

=tanA\big(\frac{(1-secA)-(1+secA)}{(1+secA)(1-secA)}\big)=tanA((1+secA)(1−secA)(1−secA)−(1+secA))

=tanA\big(\frac{(1-secA-1-secA)}{(1^{2}-sec^{2}A)}\big)=tanA((12−sec2A)(1−secA−1−secA))

=tanA\big(\frac{(-2secA)}{-(sec^{2}A-1)}\big)=tanA(−(sec2A−1)(−2secA))

=tanA\big(\frac{(-2secA)}{-tan^{2}A}\big)=tanA(−tan2A(−2secA))

/* By Trigonometric identity:

sec²A-1 = tan²A */

=\big(\frac{(2secA)}{tanA}\big)=(tanA(2secA))

=\frac{\frac{2}{cosA}}{\frac{sinA}{cosA}}=cosAsinAcosA2

=\frac{2}{sinA}=sinA2

=2cosecA=2cosecA

=RHS=RHS