Question

sin theta/ 1+ cos theta + 1+cos theta / sin theta = 2 cosec theta

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Answer 1

Answer:

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Answer 2

To Prove:-

• $\sf \dfrac{sin\theta}{1 + cos\theta} + \dfrac{1 + cos\theta}{sin\theta} = 2cosec\theta}$

SoluTion:-

$\small\sf\;\;\star\;{\underline{Taking\;L.H.S:-}}$

$\sf\maltese\; \dfrac{sin\theta}{1 + cos\theta} + \dfrac{1 + cos\theta}{sin\theta} \\ \\ \\ :\implies\sf \dfrac{sin\theta \times sin\theta + (1 + cos\theta)(1 + cos\theta)}{sin\theta(1 + cos\theta)} \\ \\ \\ :\implies\sf \dfrac{sin^2\theta + (1 + cos\theta)^2}{sin\theta(1 + cos\theta)} \\ \\ \\ :\implies\sf \dfrac{sin^2\theta + (1)^2 + 2 \times cos\theta \times 1 + (cos\theta)^2} \\ \\ \\ :\implies\sf \dfrac{sin^2\theta + (1)^2 + 2 \times cos\theta \times 1 + (cos\theta)^2}{sin\theta(1 + cos\theta)} \\ \\ \\ :\implies\sf \dfrac{sin^2\theta + 1 + 2cos\theta + cos^2\theta}{sin\theta(1 + cos\theta)}\qquad\bigg\lgroup\bf sin^2\theta + cos^2\theta = 1\bigg\rgroup$

$\\ \\ \\ :\implies\sf \dfrac{1 + 1 + 2cos\theta}{sin\theta(1 + cos\theta)} \\ \\ \\ :\implies\sf \dfrac{2 + 2cos\theta}{sin\theta(1 + cos\theta)}\qquad\bigg\lgroup\bf Taking\;2\;common \bigg\rgroup \\ \\ \\ :\implies\sf \dfrac{2 \cancel{(1 + cos\theta)}}{sin\theta \cancel{(1 + cos\theta)}} \\ \\ \\ :\implies\sf \dfrac{2}{sin\theta} \\ \\ \\ :\implies{\underline{\boxed{\bf{\blue{2cosec\theta}}}}}\;\;\bigstar$

$\dag\;{\underline{\underline{\bf{\pink{Hence\;ProvEd!!}}}}}$

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$\boxed{\begin{minipage}{7 cm} Fundamental Trigonometric Identities \\ \\ \sin^2\theta + \cos^2\theta=1 \\ \\ 1+\tan^2\theta = \sec^2\theta \\ \\ 1+\cot^2\theta = \text{cosec}^2 \, \theta \end{minipage}}$