Question

proof the following.

Answer 1

$\sf \underline{Solution : - }$

$\sf{: \implies( \tan \theta + \cot \theta + 1)(\tan \theta + \cot \theta - 1) = \sec ^{2} \theta + { \cot}^{2} \theta}$

» Consider LHS

$\sf{: \implies \big[ ( \tan \theta + \cot \theta) + 1 \big] \big[(\tan \theta + \cot \theta )- 1 \big]}$

» Apply formula (A+B)(A-B)=A²-B²

$\sf{: \implies \big[(\tan \theta + \cot \theta )^{2} - 1 \big]}$

» Apply formula (A+B)²=A²+B²+2AB

$\sf{: \implies (\tan^{2} \theta + \cot ^{2} + 2 \tan \theta.\cot \theta- 1) }$

» Apply formula tan θ . cot θ = 1

$\sf{: \implies (\tan^{2} \theta + \cot ^{2} + 2(1) - 1) }$

$\sf{: \implies (\tan^{2} \theta + \cot ^{2} + 2- 1) }$

$\sf{: \implies (\tan^{2} \theta + \cot ^{2} + 1) }$

» Apply formula tan² θ + 1 = sec² θ

$\purple{ \sf{: \implies (\sec^{2} \theta + \cot ^{2}\theta ) }}$

LHS = RHS

Hence proved