Question

Prove that root (sec^2 theta + cosec^2 theta ) = cot theta + tan theta​

Answer 1

Step-by-step explanation:

sec2theta=tan2theta+1

cosec2theta=cot2theta+1

on adding these equations, we get

sec2theta+cosec2theta=..tan2theta+cot2theta+2

==>sec2theta+cos2theta=tan2theta+cot2theta+2tanthetacotheta= (tantheta+cottheta)sq

√secsq.theta+cosecsq. theta=tantheta+cottheta .

Hence proved

Answer 2

To Prove:

• $\sf \sqrt{sec^2 \theta + cosec^2 \theta} = cot \theta + tan \theta$

⠀⠀⠀

Proof:

${\underline{\sf{\bigstar\;Taking\;LHS\;:}}}$

$:\implies\sf \sqrt{sec^2 \theta + cosec^2 \theta}$

$:\implies\sf \sqrt{ \dfrac{1}{cos^2 \theta} + \dfrac{1}{sin^2 \theta}}\qquad\qquad\bigg\lgroup\bf \because\;sec^2 \theta = \dfrac{1}{cos^2 \theta}\;,\; cosec^2 \theta = \dfrac{1}{sin^2 \theta}\bigg\rgroup$

$:\implies\sf \sqrt{ \dfrac{sin^2 \theta + cos^2 \theta}{cos^2 \theta\; sin^2 \theta}}$

$:\implies\sf \sqrt{ \dfrac{1}{cos^2 \theta \; sin^2 \theta}}\qquad\qquad\bigg\lgroup\bf \because\; sin^2 \theta + cos^2 \theta = 1\bigg\rgroup$

$:\implies\sf \dfrac{1}{cos\theta \; sin\theta}$

$:\implies\sf \dfrac{cos^2 \theta + sin^2 \theta}{cos\theta \; sin\theta}$

$:\implies\sf \dfrac{cos\; \theta}{sin \;\theta} + \dfrac{sin\; \theta}{cos \;\theta}$

$:\implies{\boxed{\sf{\purple{cot \;\theta + tan\; \theta}}}}$

$:\implies\sf LHS = RHS$

$\dag\;{\underline{\underline{\bf{Hence\;Proved!!}}}}$

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Or,

${\underline{\sf{\bigstar\;Taking\;LHS\;:}}}$

$:\implies\sf \sqrt{sec^2 \theta + cosec^2 \theta}$

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

$:\implies\sf \sqrt{2 + tan^2 \theta + cot^2 \theta}$

$:\implies\sf \sqrt{ tan^2 \theta + cot^2 \theta + 2 tan \theta cot \theta}$

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

$:\implies{\boxed{\sf{\pink{tan \;\theta + cot\; \theta}}}}$

$:\implies\sf LHS = RHS$

$\dag\;{\underline{\underline{\bf{Hence\;Proved!!}}}}$