Question

please solve this and first ans will be the brainliest ans​

Answer 1

$\large\text{\underline{Let's begin:-}}$

We have learned that $\sec\theta=\dfrac{1}{\cos\theta}$ and $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$.

$\large\text{\underline{To find:-}}$

The value of $\sin\theta$.

$\large\text{\underline{Solution:-}}$

$\dfrac{\sec\theta-\tan\theta}{\sec\theta+\tan\theta}=\dfrac{1}{4}$

$\hookrightarrow \dfrac{\dfrac{1}{\cos\theta}-\dfrac{\sin\theta}{\cos\theta}}{\dfrac{1}{\cos\theta}+\dfrac{\sin\theta}{\cos\theta}}=\dfrac{1}{4}$

Simply multiply by 1, which is $1=\dfrac{\cos\theta}{\cos\theta}$.

$\hookrightarrow \dfrac{1-\sin\theta}{1+\sin\theta}=\dfrac{1}{4}$

Solving the equation we get,

$\hookrightarrow 4(1-\sin\theta)=1+\sin\theta$

$\hookrightarrow 4-4\sin\theta=1+\sin\theta$

$\hookrightarrow 5\sin\theta=3\implies\therefore\sin\theta=\dfrac{3}{5}$

which is the wanted value.

$\large\text{\underline{Conclusion:-}}$

The required value is $\sin\theta=\dfrac{3}{5}$.

Answer 2

[ Note: Kindly see my answer from brainly app. ]

${\orange{ \boxed{ \boxed{ \begin{array}{cc} \bf \to \: given \: : \\ \\ \rm \: \frac{sec \: \theta - tan \: \theta}{sec \: \theta + tan \: \theta} = \frac{1}{4} \\ \\ \blue{ \underline{ \sf \: we \: have \: to \: find \: : }} \\ \\ \sf \: the \: value \: of \: \boxed{ \rm \: sin \: \theta \: } \\ \\ \\ \: \: \: \: \: \: \: \:\red{ \underline{ \sf \: solution \: }} \: \: \: \: \: \: \: \\ \\ \rm \frac{sec \: \theta - tan \: \theta}{sec\theta + tan \: \theta} = \frac{1}{4} \\ \\ \: \:\underbrace{{\pink{ { \boxed{ \begin{array}{cc} \sf \: we \: know \: that \: : \\ \\ \sf \hookrightarrow \: \rm \: sec \: \theta = \frac{1}{cos \: \theta} \\ \\ \sf \hookrightarrow \: \rm \:tan \: \theta = \frac{sin \: \theta}{cos \: \theta} \end{array}}}}}}_{ \underbrace{ \sf \: apply \: here}_{\downarrow}} \\ \\ \rm \implies \: \frac{ \frac{1}{cos \: \theta} - \frac{sin \: \theta}{cos \: \theta} }{ \frac{1}{cos \: \theta} + \frac{sin \: \theta}{cos \: \theta} } = \frac{1}{4} \: \: \: \: \\ \\ \rm \implies \: \frac{ \frac{1 - sin \: \theta}{cos \: \theta} }{ \frac{1 + sin \: \theta}{ cos \: \theta}} = \frac{1}{4} \\ \\ \rm \implies \:\frac{1 - sin \: \theta}{ \cancel{cos \: \theta}} \times \frac{ \cancel{ cos \: \theta}}{1 + sin \: \theta} = \frac{1}{4} \\ \\ \rm \implies \: \frac{1 - sin \: \theta}{1 + sin \: \theta} = \frac{1}{4} \\ \\ \rm \implies \:4(1 - sin \: \theta) = 1(1 + sin \: \theta) \\ \\ \rm \implies \:4 - 4sin \: \theta = 1 + sin \: \theta \\ \\ \rm \implies \: - 4sin \: \theta - sin \: \theta = 1 - 4 \\ \\ \rm \implies \: \cancel{ - } \: 5sin \: \theta = \cancel{- } \: 3 \\ \\ \rm \implies \:5sin \: \theta = 3 \\ \\ \rm \implies \:sin \: \theta = \frac{3}{5} \\ \\ \\ \blue{ \boxed{\therefore \rm \: sin \: \theta = \frac{3}{5} }}\end{array}}}}}$