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Maharashtra State BoardSSC (English Medium) 10th Standard

Prove that (cot A)/(1 – tan A) + (tan A)/(1 – cot A) = 1 + tan A + cot A = sec A . cosec A + 1.

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Question

Prove that `(cot A)/(1 - tan A) + (tan A)/(1 - cot A) = 1 + tan A + cot A = sec A  .  "cosec"  A + 1`.

Theorem
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Solution

`(cot A)/(1 - tan A) + (tan A)/(1 - cot A)`

= `((cos A)/(sin A))/(1 - (sin A)/(cos A)) + ((sin A)/(cos A))/(1 - (cos A)/(sin A))`

= `((cos A)/(sin A))/((cos A  -  sin A)/(cos A)) + ((sin A)/(cos A))/((sin A  -  cos A)/(sin A))`

= `(cos A)/(sin A) xx (cos A)/(cos A - sin A) + (sin A)/(cos A) xx (sin A)/(sin A - cos A)`

= `(cos^2A)/(sin A(cos A - sin A)) + (sin^2A)/(cos A(sin A - cos A))`

= `1/(sin A - cos A) ((-cos^3A + sin^3A)/(sin A cos A))`

= `1/(sin A - cos A)((sin^3A - cos^3A)/(sin A cos A))`

= `1/(sin A - cos A) xx ((sin A - cos A)(sin^2A + sin A cos A + cos^2A))/(sin A cos A)`   ...[∵ a3 – b3 = (a – b)(a2 + ab + b2)]

= `(sin^2A + sin A cos A + cos^2A)/(sin A cos A)`   ...(i)

= `(1 + sin A cos A)/(sin A cos A)`   ...[∵ sin2A + cos2A = 1]

= `1/(sin A cos A) + (sin A cos A)/(sin A cos A)`

= cosec A sec A + 1   ...(ii)

`(cot A)/(1 - tan A) + (tan A)/(1 - cot A)`

= `(sin^2A + sin A cos A + cos^2A)/(sin A cos A)`   ...[From (i)]

= `(sin^2A)/(sin A cos A) + (sin A cos A)/(sin A cos A) + (cos^2A)/(sin A cos A)`

= `(sin A)/(cos A) + 1 + (cos A)/(sin A)`

= tan A + 1 + cot A   ...(iii)

From (ii) and (iii), we get

`(cot A)/(1 - tan A) + (tan A)/(1 - cot A) = 1 + tan A + cot A = sec A  .  "cosec"  A + 1`

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Chapter 6: Trigonometry - Exercise

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