Question

cos theta is equals to under root 3 upon 2 then find the value of 1 minus sin theta upon 1 + sec theta ​

Answer 1

$\underline{\textsf{Given:}}$

$\mathsf{cos\theta=\dfrac{\sqrt{3}}{2}}$

$\underline{\textsf{To find:}}$

$\textsf{The value of}\;\mathsf{\dfrac{1-sin\theta}{1+sec\theta}}$

$\underline{\textsf{Solution:}}$

$\textsf{Consider,}$

$\mathsf{cos\theta=\dfrac{\sqrt{3}}{2}}$

$\implies\mathsf{\theta=30^{\circ}}$

$\textsf{Then,}$

$\mathsf{sin\,\theta=sin\,30^{\circ}=\dfrac{1}{2}}$

$\textsf{Now}$

$\mathsf{\dfrac{1-sin\theta}{1+sec\theta}}$

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

$\mathsf{=\dfrac{1-\dfrac{1}{2}}{1+\dfrac{2}{\sqrt{3}}}}$

$\mathsf{=\dfrac{\dfrac{1}{2}}{1+\dfrac{2}{\sqrt{3}}}}}$

$\mathsf{=\dfrac{\dfrac{1}{2}}{\dfrac{\sqrt{3}+2}{\sqrt{3}}}}$

$\mathsf{=\dfrac{\sqrt{3}}{2(2+\sqrt{3})}}$

$\textsf{To rationalize the denominator}$

$\mathsf{=\dfrac{\sqrt{3}}{2(2+\sqrt{3})}{\times}\dfrac{2-\sqrt{3}}{2-\sqrt{3}}}$

$\mathsf{=\dfrac{\sqrt{3}(2-\sqrt{3})}{2^2-\sqrt{3}^2}}$

$\mathsf{=\dfrac{\sqrt{3}(2-\sqrt{3})}{4-3}}$

$\mathsf{=\dfrac{\sqrt{3}(2-\sqrt{3})}{1}}$

$\mathsf{=2\sqrt{3}-3}$

$\underline{\textsf{Answer:}}$

$\textsf{The value of}\;\mathsf{\dfrac{1-sin\theta}{1+sec\theta}\;\textsf{is}\;\mathsf{2\sqrt{3}-3}}$

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