Question

Prove that [sec -tan]2 =1-sin/1+sin

Answer 1

here's your answer. mark brainliest if I was able to clear your doubt

Answer 2

To prove that: $(\sec A -\tan A)^{2}$ = $\dfrac{1-\sin A}{1+\sin A}$.

Solution:

R.H.S. = $\dfrac{1-\sin A}{1+\sin A}$

Rationalising numerator and denominator, we get

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

= $\dfrac{(1-\sin A)^2}{1^2-\sin^2 A}$

Using the algebraic identity:

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

= $\dfrac{1+\sin^2 A-2\sin A}{1-\sin^2 A}$ [ ∵ $(a-b)^{2} =a^{2} +b^{2}$ - 2ab]

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

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

= $\sec^2 A+\tan^2 A-2\sec A\tan A$

= $(\sec A-\tan A)^2$

= L.H.S., proved.

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