Prove that √sec2θ + cosec2θ = tan θ + cot θ .

1.	Prove that √sec2θ + cosec2θ = tan θ + cot θ .
Answer» TO PROVE :– $\\ \implies { \bold{ \sqrt{ { \sec}^{2} \theta + {cosec}^{2} \theta } = \tan( \theta) + \cot( \theta) }} \\$ SOLUTION :– • Let's TAKE L.H.S. – $\\ \: \: = \: \: { \bold{ \sqrt{ { \sec}^{2} \theta + {cosec}^{2} \theta }}} \\$ ▪︎ Using identity – $\\ \: \: \dashrightarrow\: \: { \bold{ { \sec}^{2} \theta = 1 + { \tan}^{2} \theta}} \\$ $\\ \: \: \dashrightarrow\: \: { \bold{ { cosec}^{2} \theta = 1 + { \cot}^{2} \theta}} \\$ • So that – $\\ \: \: = \: \: { \bold{ \sqrt{1 + { \tan}^{2} \theta+1 + { \cot}^{2} \theta}}} \\$ $\\ \: \: = \: \: { \bold{ \sqrt{ { \tan}^{2} \theta+2+ { \cot}^{2} \theta}}} \\$ • We should WRITE this as – $\\ \: \: = \: \: { \bold{ \sqrt{ { \tan}^{2} \theta+2 \tan \theta \cot \theta + { \cot}^{2} \theta}}} \\$ • Using identity – $\\ \: \: \dashrightarrow\: \: { \bold{ {a}^{2} + 2ab + {b}^{2} = {(a + b)}^{2} }} \\$ $\\ \: \: = \: \: { \bold{ \sqrt{{( \tan \theta + \cot \theta )}^{2}}}} \\$ $\\ \: \: = \: \: { \bold{ \tan \theta + \cot \theta }} \\$ $\\ \: \: = \: \: { \bold{R.H.S. }} \\$ $\\ \: \: \: \: \: { \underbrace{ \bold{Hence \: \: proved }}} \\$

Answer»

TO PROVE :–

$\\ \implies { \bold{ \sqrt{ { \sec}^{2} \theta + {cosec}^{2} \theta } = \tan( \theta) + \cot( \theta) }} \\$

SOLUTION :–

• Let's TAKE L.H.S. –

$\\ \: \: = \: \: { \bold{ \sqrt{ { \sec}^{2} \theta + {cosec}^{2} \theta }}} \\$

▪︎ Using identity –

$\\ \: \: \dashrightarrow\: \: { \bold{ { \sec}^{2} \theta = 1 + { \tan}^{2} \theta}} \\$

$\\ \: \: \dashrightarrow\: \: { \bold{ { cosec}^{2} \theta = 1 + { \cot}^{2} \theta}} \\$

• So that –

$\\ \: \: = \: \: { \bold{ \sqrt{1 + { \tan}^{2} \theta+1 + { \cot}^{2} \theta}}} \\$

$\\ \: \: = \: \: { \bold{ \sqrt{ { \tan}^{2} \theta+2+ { \cot}^{2} \theta}}} \\$

• We should WRITE this as –

$\\ \: \: = \: \: { \bold{ \sqrt{ { \tan}^{2} \theta+2 \tan \theta \cot \theta + { \cot}^{2} \theta}}} \\$

• Using identity –

$\\ \: \: \dashrightarrow\: \: { \bold{ {a}^{2} + 2ab + {b}^{2} = {(a + b)}^{2} }} \\$

$\\ \: \: = \: \: { \bold{ \sqrt{{( \tan \theta + \cot \theta )}^{2}}}} \\$

$\\ \: \: = \: \: { \bold{ \tan \theta + \cot \theta }} \\$

$\\ \: \: = \: \: { \bold{R.H.S. }} \\$

$\\ \: \: \: \: \: { \underbrace{ \bold{Hence \: \: proved }}} \\$

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