Question

show that :- cosec square theta - tan square ( 90 - theta) = sin square theta + sin square ( 90 - theta

Answer 1

Answer:

In the attachment.. Hope it's helpful..

Answer 2

Answer:

Step-by-step explanation:

To prove : cosec^2 θ - tan^2 ( 90 -  θ ) = sin^2  θ + sin^2( 90 -  θ )

Solving left hand side,

= > cosec^2 θ - tan^2 ( 90 -  θ )

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From the properties of trigonometry, we know : -

tan^2 ( 90 - A ) = cot^2 A

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= > cosec^2 θ - cot^2 A

$\implies \dfrac{\text{hypotenuse}^2}{\text{height}^2}-\dfrac{\text{Base}^2}{\text{height}^2}\\\\\\\implies\dfrac{\text{hypotenuse}^2-\text{base}^2}{\text{height}^2}\\\\\\\implies\dfrac{\text{height}^2}{\text{height}^2}\\\\\\\implies 1$

Thus, left hand side = > 1

Solving right hand side,

$\implies \dfrac{\text{height}^2}{\text{hypotenuse}^2}+\dfrac{\text{Base}^2}{\text{hypotenuse}^2}\\\\\\\implies\dfrac{\text{height}^2+\text{base}^2}{\text{hypotenuse}^2}\\\\\\\implies\dfrac{\text{hypotenuse}^2}{\text{hypotenuse}^2}\\\\\\\implies 1$

Thus, right hand side = > 1

1 = 1

Left hand side = Right hand side

Hence, proved.