Question

(1+cot²thetha) sinthetha=1

Answer 1

Answer:

cosec

$( {cot}^{2}thetha + 1) \times \ {sin}^{2} \: theha = {cosec}^{2} \times \frac{1}{c {cosec}^{2} } = 1$

tou have written wrong question

because you have written sinthetha it should be sin square thetha

Answer 2

Answer

$\bullet \;\; \rm (1+cot^2\theta)sin^2\theta=1$

Given

$\bullet \;\; \rm (1+cot^2\theta)sin^2\theta$

To prove

$\bullet \;\; \rm (1+cot^2\theta)sin^2\theta=1$

Proof

$\rm LHS\\\\\implies \rm (1+cot^2\theta)sin^2\theta\\\\\implies \rm \bigg(1+\dfrac{cos^2\theta}{sin^2\theta}\bigg)sin^2\theta\\\\\implies \rm \bigg(\dfrac{sin^2\theta+cos^2\theta}{sin^2\theta}\bigg)sin^2\theta\\\\\implies \rm \bigg(\dfrac{sin^2\theta+cos^2\theta}{\cancel{sin^2\theta}}\bigg)\cancel{sin^2\theta}\\\\\implies \rm sin^2\theta+cos^2\theta\\\\\implies \rm 1\\\\\implies \rm RHS\\\\\rm Hence\ proved$

Explore More

$\bullet \bf \;\; cot\theta=\dfrac{cos\theta}{sin\theta}\\\\\bullet \bf \;\; sin^2\theta+cos^2\theta=1$

$\boxed{\begin{minipage}{5cm} \underbrace{\bf Trigonometric\ Identities}\\\\\bullet \;\; \rm sin^2\theta+cos^2\theta=1\\\\\bullet \;\; \rm sec^2\theta-tan^2\theta=1\\\\\bullet \;\; \rm cosec^2\theta-cot^2\theta=1 \end{minipage}}$