Question

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

Answer 1

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 2

Step-by-step explanation:

${\red{\underline{\underline{\bold{To\:Prove:-}}}}}$

$\sqrt{ \sec2 \theta + \cosec2 \theta } = \tan \theta + \cot \theta$

${\green{\underline{\underline{\bold{Proof:-}}}}}$

$\sqrt{ \sec2 \theta + \cosec2 \theta } = \tan \theta + \cot \theta$

As we know that:-

$\sec \theta = \frac{1}{ \cos \theta } \\\\</p><p></p><p> \cosec \theta = \frac{1}{ \sin \theta }$

$L.H.S = \sqrt{ \sec2 \theta + \cosec2 \theta }\\\\ </p><p></p><p>= \sqrt{ \frac{1}{ { \cos }^{2} \theta} + \frac{1}{ { \sin }^{2} \theta} } \\\\</p><p></p><p>= \sqrt{ \frac{ { \sin }^{2} \theta + { \cos}^{2} \theta }{ { (\sin }^{2} \theta) ( { \cos }^{2} \theta)} } \\\\</p><p></p><p>= \sqrt{ \frac{ 1 }{ { \sin }^{2} \theta. { \cos }^{2} \theta} } \: \: \: ( \therefore { \sin }^{2} \theta + { \cos }^{2} \theta = 1) \\\\ </p><p></p><p>= \frac{1}{ \sin \theta. \cos \theta } \\\\</p><p></p><p>= \frac{ { \sin }^{2} \theta + { \cos }^{2} \theta }{ \sin \theta. \cos \theta } \\\\</p><p></p><p>= \frac{ \sin \theta. \sin \theta }{ \sin\theta. \cos \theta} + \frac{ \cos \theta. \cos \theta }{ \sin\theta. \cos \theta} \\\\</p><p></p><p>= \frac{ \sin \theta }{ \cos \theta} + \frac{ \cos \theta }{ \sin\theta} \\\\</p><p></p><p>= \tan \theta + \cot \theta$

${\purple{\underline{\underline{\bold{Hence\:Proved}}}}}$