Question

prove that cos theta +1- sin theta upon cos theta - 1+ sin theta =1+ cos theta upon sin theta
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Answer 1

Given :-

$\bullet\sf \ \dfrac {cos\theta+1-sin\theta}{cos\theta-1+sin\theta}= \dfrac{1+cos\theta}{sin\theta}$

Solution :-

LHS :-

$:\implies\sf\ \dfrac{^{(cos\theta+1-sin\theta)}/_{sin\theta}}{^{(cos\theta-1+sin\theta)}/_{sin\theta}}\ \ \ \ \ \ \big\lgroup\ Dividing\ By\ sin\theta \big\rgroup\\ \\ \\ :\implies\sf \ \dfrac{cot\theta+csc\theta-1}{cot\theta-csc\theta+1}\\ \\ \\ :\implies\sf \ \dfrac{cot\theta+csc\theta-\big(csc^2\theta-cot^2\theta\big)}{cot\theta-csc\theta+1}\ \ \ \ \ \ \big\lgroup\ 1= csc^2\theta-cot^2\theta \big\rgroup$

$\\ \\ \\ :\implies\sf\ \dfrac{csc\theta+cot\theta-\big(csc\theta+cot\theta\big)\big(csc\theta-cos\theta \big)}{cot\theta-csc\theta+1}\ \ \ \ \ \big\lgroup\ a^2-b^2=(a+b)(a-b) \big\rgroup\\ \\ \\ :\implies\sf \ \dfrac{csc\theta+cos\theta \{1-(csc\theta-cot\theta)\}}{cot\theta-csc\theta+1}\\ \\ \\ :\implies\sf \ \dfrac{csc\theta+cot\theta \cancel{\{1-csc\theta+cot\theta\}}}{\cancel{cot\theta-csc\theta+1}}\\ \\ \\ :\implies\sf\ csc\theta+cot\theta\\ \\ \\ :\implies\sf \ \dfrac{1}{sin\theta}+\dfrac{cos\theta}{sin\theta}\\ \\ \\ :\implies\underline{\boxed{\purple{\sf \dfrac{1+cos\theta}{sin\theta}}}}\ \ \ \ \ \sf \ Proved !$

Answer 2

Answer:

$\bf{To\:Prove :}\:\sf\dfrac{cos\theta+1-sin\theta}{cos\theta-1+sin\theta}=\dfrac{1+sin\theta}{cos\theta}$

$\bf{Proof :}$

$:\implies\sf\ \dfrac{ \frac{(cos\theta+1-sin\theta)}{sin\theta}}{ \frac{(cos\theta-1+sin\theta)}{sin\theta}}\\\\{\scriptsize\qquad\bf{\dag}\:\:\tt{Dividing\ By\ sin\theta}}\\ \\ :\implies\sf \ \dfrac{cot\theta+csc\theta-1}{cot\theta-csc\theta+1}\\ \\ \\ :\implies\sf \ \dfrac{cot\theta+csc\theta-\big(csc^2\theta-cot^2\theta\big)}{cot\theta-csc\theta+1}\\ \\ {\scriptsize\qquad\bf{\dag}\:\:\tt{1= csc^2\theta-cot^2\theta}}$

⠀

$:\implies\sf\ \dfrac{csc\theta+cot\theta-\big(csc\theta+cot\theta\big)\big(csc\theta-cos\theta \big)}{cot\theta-csc\theta+1}\\ \\ {\scriptsize\qquad\bf{\dag}\:\:\tt{\ a^2-b^2=(a+b)(a-b)}} \\ \\ :\implies\sf \ \dfrac{csc\theta+cos\theta [1-(csc\theta-cot\theta)]}{cot\theta-csc\theta+1}\\ \\ \\ :\implies\sf \ \dfrac{csc\theta+cot\theta \bcancel{[1-csc\theta+cot\theta]}}{\bcancel{cot\theta-csc\theta+1}}\\ \\ \\ :\implies\sf\ csc\theta+cot\theta\\ \\ \\ :\implies\sf \ \dfrac{1}{sin\theta}+\dfrac{cos\theta}{sin\theta}\\ \\ \\ :\implies\underline{\boxed{\sf \dfrac{1+cos\theta}{sin\theta}}}$