Question

7. If sec 0 - tan 0 / sec 0 + tan 0 = 1/4, find sin 0​

Answer 1

$\underline{\underline{\sf SOLUTION :}}$

$\dashrightarrow\:\:\sf \dfrac{\sec( \theta) - \tan( \theta)}{\sec( \theta) + \tan( \theta)} = \dfrac{1}{4}$

$\tiny\dag \: \underline{\frak{Cross \: multiplying \: both \: the \: sides \: we \: get :}}$

$\dashrightarrow\:\:\sf 4 \bigg(\sec( \theta) - \tan( \theta) \bigg) = 1 \bigg(\sec( \theta) + \tan( \theta) \bigg)$

$\dashrightarrow\:\:\sf 4 \sec( \theta) - 4 \tan( \theta) = \sec( \theta) + \tan( \theta)$

$\tiny\dag \: \underline{\frak{Now \: Combine\: the \: like \:terms\:we \: get :}}$

$\dashrightarrow\:\:\sf 4 \sec( \theta) -\sec( \theta) = \tan( \theta) + 4 \tan( \theta)$

$\dashrightarrow\:\:\sf 3 \sec( \theta) = 5 \tan( \theta)$

$\dashrightarrow\:\:\sf 3 \sec( \theta) = 5 \times \dfrac{ \sec( \theta)}{ \csc(\theta) }\qquad \Bigg\lgroup\bf{\because \tan(\theta) = \dfrac{\sec( \theta)}{\csc(\theta)} }\Bigg\rgroup$

$\dashrightarrow\:\:\sf \dfrac{ 3 \sec( \theta)}{5\sec( \theta)} = \dfrac{1}{ \csc(\theta) }$

$\dashrightarrow\:\:\sf \dfrac{ 3 }{5} = \dfrac{1}{ \csc(\theta) }$

$\tiny\dag \: \underline{\frak{Replacing \dfrac{1}{\csc (\theta)} by \sin(\theta):}}$

$\dashrightarrow\:\: \dag \: \underline{ \boxed{\sf \sin( \theta) = \dfrac{ 3 }{5} }} \: \dag$

Answer 2

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