But What If sin ⁡ A cos ⁡ A = 0 \sin A \cos A = 0 ?

Geometry Level 3

( 1 + cot ⁡ A − csc ⁡ A ) ( 1 + tan ⁡ A + sec ⁡ A ) \large (1 + \cot A - \csc A )(1 + \tan A + \sec A )

If A A can be any angle such that sin ⁡ A cos ⁡ A ≠ 0 \sin A \cos A \ne 0 , simplify the trigonometric expression above.

-2 -1 0 1 2

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Ashrit Ramadurgam by Ashrit Ramadurgam , 20, India

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1 solution

Ashrit Ramadurgam
Mar 25, 2016

( 1 + cot ⁡ A − csc ⁡ A ) ( 1 + tan ⁡ A + sec ⁡ A ) (1 + \cot A - \csc A)(1 + \tan A + \sec A) = ( 1 + cos ⁡ A sin ⁡ A − 1 sin ⁡ A ) ( 1 + sin ⁡ A cos ⁡ A + 1 cos ⁡ A ) =\Big( 1 + \frac{\cos A}{\sin A} - \frac{1}{\sin A} \Big) \Big( 1 + \frac{\sin A}{\cos A} + \frac{1}{\cos A} \Big) = ( sin ⁡ A + cos ⁡ A − 1 sin ⁡ A ) ( sin ⁡ A + cos ⁡ A + 1 cos ⁡ A ) =\Big( \frac{\sin A + \cos A - 1}{\sin A} \Big) \Big( \frac{\sin A + \cos A + 1}{\cos A} \Big) = ( sin ⁡ A + cos ⁡ A ) 2 − 1 2 sin ⁡ A cos ⁡ A =\frac{(\sin A + \cos A)^2 - 1^2}{\sin A \cos A} = sin ⁡ 2 A + cos ⁡ 2 A + 2 sin ⁡ A cos ⁡ A − 1 sin ⁡ A cos ⁡ A =\frac{\sin ^2A + \cos ^2A + 2 \sin A \cos A - 1}{\sin A \cos A} = 1 + 2 sin ⁡ A cos ⁡ A − 1 sin ⁡ A cos ⁡ A ⋯ [ ∵ sin ⁡ 2 A + cos ⁡ 2 A = 1 ] =\frac{1 + 2 \sin A \cos A - 1}{\sin A \cos A} \cdots \big[ \because \sin ^2A + \cos ^2A = 1 \big] = 2 sin ⁡ A cos ⁡ A sin ⁡ A cos ⁡ A =\frac{2 \sin A \cos A}{\sin A \cos A} = 2 ⋯ [ ∵ sin ⁡ A cos ⁡ A ≠ 0 ] =\boxed{2} \cdots \big[ \because \sin A \cos A \ne 0 \big]

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