Question

Prove that(1+ tan A-sec A)x (1+tan A + sec A) =2tanA
​

Answer 1

To Prove :-

$\sf (1+tanA-secA) \times (1+tanA+secA) = 2tanA$

Solution :-

$\sf L.H.S = (1+tanA-secA)\times (1+tanA+secA)$

       $= \sf ((1+tanA)-secA) \times ((1+tanA) +secA)$

Using the identity ( a- b ) ( a + b ) = a² - b²

        $= \sf (1+tan^2A)-sec^2A \\\\= 1+tan^2A+2tanA-sec^2A$

$\bullet \bf \ 1+tan^2A= sec^2A$

        $=\sf sec^2A+2tanA-sec^2A \\\\= 2tanA+sec^2A-sec^2A \\\\= 2tanA\\\\= R.H.S$

Hence proved.

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$