Question

Prove that sec A (1 – sin A)(sec A + tan A) = 1.​

Answer 1

Given:-

$\odot \sf sec A ( 1 - sin A) (sec A + tanA) = 1$

To find:-

Here we have to prove the R.H.S side.

Solution:-

$\odot \sf sec A ( 1 - sin A) (sec A + tan A) = 1$

$\red\bigstar$Firstly write everything in term of sin A and cos A.

$\implies \sf \dfrac{1}{cos A} (1 - sin A) \bigg(\dfrac{1}{cos A} + \dfrac{sin A}{cos A}\bigg) \\ \\\bigg(\therefore \sf sec A = \dfrac{1}{cos A} \: and \: Tan A = \dfrac{sin A}{cos A}\bigg) \\ \\ \\ \implies \sf \dfrac{(1-sin A)}{cos A} \: \bigg( \dfrac{1 + sin A}{cos A}\bigg) \\ \\ \\ \implies \sf = \dfrac{(1 - sin A) \: (1 + sin A)}{cos A \times cos A}$

$\red\bigstar$We know that (a - b) (a + b) = a² + b²

$\implies \sf \dfrac{(1^{2} - sin^{2} A)}{cos^{2}A} \\ \\ \implies \sf \dfrac{(1 - sin^{2} A)}{cos^{2} A} \\ \\ \implies \sf \dfrac{cos^{2}A}{cos^{2}A} \: \: \sf (\therefore cos^{2} A + sin^{2} A = 1) \\ \\ \implies = 1 \\ \\ \implies \sf = R.H.S \: Hence, \: proved$

Additional:-

Some identities to know:-

$\sf \diamondsuit sin^{2}\theta + cos^{2}\theta = 1 \\ \diamondsuit \sf sin^{2} \theta = 1 - cos^{2}\theta \\ \diamondsuit \sf sin\theta = \sqrt{1 - cos^{2}\theta} \\ \diamondsuit \sf cos^{2}\theta = 1 - sin^{2}\theta \\ \diamondsuit \sf cos \theta = \sqrt{1 - sin^{2}\theta} \\ \\ \\ \odot \sf Sec^{2}\theta - tan^{2}\theta = 1 \\ \odot \sf sec^{2}\theta = 1 + tan^{2}\theta \\ \odot \sf sec\theta = \sqrt{ 1+ tan^{2}\theta} \\ \odot \sf tan^{2}\theta = sec^{2}\theta - 1 \\ \odot \sf tan\theta = \sqrt{sec^{2} \theta - 1} \\ \\ \\ \implies \sf cosec^{2}\theta - cot^{2}\theta = 1 \\ \implies \sf cosec^{2}\theta= 1 + cot^{2}\theta \\ \implies \sf cosec\theta = \sqrt{1 + cot^{2}\theta} \\ \implies \sf cot^{2}\theta = cosec^{2}\theta - 1 \\ \implies \sf cot\theta = \sqrt{cosec^{2}\theta - 1}$