Question

Please answer the questions in the image. Thanks in advance​

Answer 1

$\mathsf{2.\;\;\;Given : 4cos\theta - 11sin\theta = 0}$

$\mathsf{\implies 11sin\theta = 4cos\theta}$

$\mathsf{\implies \dfrac{sin\theta}{cos\theta} = \dfrac{4}{11}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{tan\theta = \dfrac{sin\theta}{cos\theta}}}}$

$\mathsf{\implies tan\theta = \dfrac{4}{11}}$

$\mathsf{Now,\;Consider : \dfrac{11cos\theta - 7sin\theta}{11cos\theta + 7sin\theta}}$

$\textsf{Dividing Numerator and Denominator of above fraction with \mathsf{cos\theta }, We get :}$

$\mathsf{\implies \dfrac{\bigg(\dfrac{11cos\theta - 7sin\theta}{cos\theta}\bigg)}{\bigg(\dfrac{11cos\theta + 7sin\theta}{cos\theta}\bigg)}}$

$\mathsf{\implies \dfrac{{\bigg(\dfrac{11cos\theta}{cos\theta} - \dfrac{7sin\theta}{cos\theta}\bigg)}}{\bigg(\dfrac{11cos\theta}{cos\theta} + \dfrac{7sin\theta}{cos\theta}\bigg)}}$

$\mathsf{\implies \dfrac{{(11 - 7tan\theta)}}{(11 + 7tan\theta)}}$

$\mathsf{But,\;We\;found\;that : tan\theta = \dfrac{4}{11}}$

$\mathsf{\implies \dfrac{11 - 7\bigg(\dfrac{4}{11}\bigg)}{11 + 7\bigg(\dfrac{4}{11}\bigg)}}$

$\mathsf{\implies \dfrac{11 - \bigg(\dfrac{28}{11}\bigg)}{11 + \bigg(\dfrac{28}{11}\bigg)}}$

$\mathsf{\implies \dfrac{\bigg(\dfrac{121 - 28}{11}\bigg)}{\bigg(\dfrac{121 + 28}{11}\bigg)}}$

$\mathsf{\implies \dfrac{93}{149}}$

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$\textsf{3.\;\;\;Given : 2sinA = 1}$

$\mathsf{\implies sinA = \dfrac{1}{2}}$

$\textsf{Now, Consider : secA - tanA}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{sec\theta = \dfrac{1}{cos\theta}}}}$

$\mathsf{\implies secA - tanA = \dfrac{1}{cosA} - \dfrac{sinA}{cosA}}$

$\mathsf{\implies secA - tanA = \dfrac{1 - sinA}{cosA}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{sin^2\theta + cos^2\theta = 1}}}$

$\mathsf{\implies cos^2\theta = 1 - sin^2\theta}$

$\mathsf{\implies cos\theta = \pm\sqrt{1 - sin^2\theta}}$

$\mathsf{\implies secA - tanA = \pm\dfrac{1 - sinA}{\sqrt{1 - sin^2A}}}$

$\mathsf{\implies secA - tanA = \pm\dfrac{1 - \dfrac{1}{2}}{\sqrt{1 - \bigg(\dfrac{1}{2}\bigg)^2}}}$

$\mathsf{\implies secA - tanA = \pm\dfrac{\dfrac{1}{2}}{\sqrt{1 - \dfrac{1}{4}}}}$

$\mathsf{\implies secA - tanA = \pm\dfrac{\dfrac{1}{2}}{\sqrt{\dfrac{4 - 1}{4}}}}$

$\mathsf{\implies secA - tanA = \pm\dfrac{\dfrac{1}{2}}{\sqrt{\dfrac{3}{4}}}}$

$\mathsf{\implies secA - tanA = \pm\dfrac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}}$

$\mathsf{\implies secA - tanA = \pm\dfrac{1}{\sqrt{3}}}$