Question

(sec- A-1)² - (tanA - sinA)² = (1-cosA)²​

Answer 1

Refer attachment for your answer...

Answer 2

To Prove :-

$\sf (secA-1)^2-(tanA-sinA)^2 = (1-cosA)^2$

Solution :-

$\sf L.H.S = (secA-1)^2- (tanA-sinA)^2= ( 1-cosA)^2$

        $= \sf sec^2A+1-2secA -(tan^2A+sin^2A-2tanAsinA ) \\\\= sec^2A+1-2secA-tan^2A-sin^2A+2tanAsinA \\\\= (sec^2A-tan^2A)+(1-sin^2A)-2secA+2tanAsinA$

$\bullet \bf \ sec^2-tan^2A= 1 \\\\\bullet \ 1-sin^2A = cos^2A$

        $= \sf 1+cos^2A-2secA+2tanAsinA \\\\= 1+cos^2A -2(secA+tanAsinA)$

$\bullet \bf \ secA= \dfrac{1}{cosA} \\\\\bullet \ tanA = \dfrac{sinA}{cosA}$

      $= \sf 1+cos^2A -2 \left( \dfrac{1}{cosA} +\dfrac{sinA}{cosA} \times sinA \right) \\\\= 1+cos^2A- 2 \left( \dfrac{1-sin^2A}{cosA} \right) \\\\= 1+cos^2A -2 \times \dfrac{cos^2A}{cosA} \\\\= 1+cos^2A - 2cosA \\\\= (1-cosA)^2\\\\= R.H.S$

Hence proved.