Question

2.
Show that (cosec 0 - cot 0)2 =
1- cos e
1+ cos 0​

Answer 1

$\huge\mathbb{ANSWER}$

$\sf\underline{\blue{\:\:\:\:\: Given:-\:\:\:\:\:}}$

$\sf \hookrightarrow(cosec\theta -cot\theta){}^{2}=\dfrac{1-cos\theta}{1+cos\theta}$

• We have to prove that

$\sf\underline{\pink{\:\:\:\:\: Solution:-\:\:\:\:\:}}$

$\sf \leadsto cosec \:\theta=\dfrac{1}{sin\theta}\\ \\ \sf \leadsto cot \:\theta=\dfrac{cos\theta}{sin\:\sin\theta}$

Now :-

• L.H.S

$\sf\implies (cosec\theta -cot\theta){}^{2}\\ \\ \sf\implies cosec{}^{2}\theta+cot{}^{2}\theta-2cot\theta.cosec\theta\\ \\ \sf\:\: put\:the\:value\:of\:cosec\theta\:and\:cot\:theta\\ \\ \sf\implies \dfrac{1}{sin{}^{2}\theta}+\dfrac{cos{}^{2}\theta}{sin{}^{2}\theta}+2\times \dfrac{1}{sin\theta}\times \dfrac{cos\theta}{sin\theta}\\ \\ \sf\implies \dfrac{1+cos{}^{2}\theta}{sin{}^{2}\theta} -\dfrac{2cos\theta}{sin{}^{2}\theta}\\ \\ \sf\implies \dfrac{1+cos{}^{2}\theta-2cos\theta}{sin{}^{2}\theta}\\ \\ \sf \:\:\:it\: become\:in\:form\:of\:(a-b){}^{2}\\ \\ \sf\implies sin{}^{2}\theta=1-cos{}^{2}\theta\\ \\ \sf\implies\dfrac{(1-cos\theta){}^{2}}{1-cos{}^{2}\theta}\\ \\ \sf\implies \dfrac{(1-cos\theta)\times \cancel{(1-cos\theta)}}{(1+cos\theta)\times \cancel{(1-cos\theta)}}\\ \\ \sf\implies \dfrac{1-cos\theta}{1+cos\theta}\\ \\ \sf\:\:\:Hence\:\: Proved !!$

$\boxed{\sf{\purple{\dfrac{1-cos\theta}{1+cos\theta}=\dfrac{1-cos\theta}{1+cos\theta}}}}$

Answer 2

$\huge{\mathfrak{\underline{\underline \green{Answer}}}}$

$\large{\sf{\underline{\underline \blue{Given:-}}}}$

•We have to prove that

$\large\tt\ (cosec \thetha \: - \: cot \thetha)^2 \: = \: \frac{1 \: - \: cos \thetha}{1 \: + \: cos \thetha}$