Question

Sin(90-a)/1+sina+cosa/1-cos(90-a)prove 2seca

Answer 1

Answer:

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Answer 2

$\dfrac{\sin(90-A)}{1+\sin A}+\dfrac{\cos A}{1-\cos(90-A)}=2\sec A$, proved.

Step-by-step explanation:

To prove that, $\dfrac{\sin(90-A)}{1+\sin A}+\dfrac{\cos A}{1-\cos(90-A)}=2\sec A$

L.H.S. = $\dfrac{\sin(90-A)}{1+\sin A}+\dfrac{\cos A}{1-\cos(90-A)}$

= $\dfrac{\cos A}{1+\sin A}+\dfrac{\cos A}{1-\sin A}$

Using the trigonometric identity,

$\cos A=\sin(90-A)$

and $\sin A=\cos(90-A)$

= $\dfrac{\cos A(1-\sin A)+\cos A(1+\sin A)}{(1+\sin A)(1-\sin A)}$

= $\dfrac{\cos A-\cos A\sin A+\cos A+\cos A\sin A}{1^2-\sin^2 A}$

Using the algebraic identity,

$a^{2}-b^{2} =(a+b)(a-b)$

= $\dfrac{\cos A+\cos A}{1-\sin^2 A}$

= $\dfrac{2\cos A}{\cos^2 A}$

Using the trigonometric identity,

$\sin^2 A+\cos^2 A=1$

= $\dfrac{2}{\cos A}$

Using the trigonometric identity,

$\sec A=\dfrac{1}{\cos A}$

= $2\sec A$

= R.H.S., proved.

Thus, $\dfrac{\sin(90-A)}{1+\sin A}+\dfrac{\cos A}{1-\cos(90-A)}=2\sec A$, proved.