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Find $\sin{\theta}$, if $\sec{\theta}-\tan{\theta} = p$

Theta is an angle of a right triangle and the difference of secant and tangent functions is equal to $p$.

Find reciprocal of given trigonometric equation

The Pythagorean identity of secant and tangent functions can be expressed in reciprocal form. Actually, the trigonometric expression $\sec{\theta}-\tan{\theta}$ is one of them and it is part of the given trigonometric equation.

$sec^2{\theta}-\tan^2{\theta} = 1$

$\implies {(\sec{\theta}+\tan{\theta})}$ ${(\sec{\theta}-\tan{\theta})}$ $=$ $1$

It is given in this trigonometry problem that $\sec{\theta}-\tan{\theta} = p$. Substitute, $\sec{\theta}-\tan{\theta}$ in this equation to get value of its reciprocal.

$\implies$ ${(\sec{\theta}+\tan{\theta})}$ $\times p$ $=$ $1$

$\implies$ $\sec{\theta}+\tan{\theta}$ $=$ $\dfrac{1}{p}$

Add trigonometric equations for each function

Add sum and difference equations of sec and tan functions and it is useful to obtain the value of tan function.

${(\sec{\theta}-\tan{\theta})}$ $+$ ${(\sec{\theta}+\tan{\theta})}$ $\,=\,$ $p+\dfrac{1}{p}$

$\implies$ $\sec{\theta}-\tan{\theta}$ $+$ $\sec{\theta}+\tan{\theta}$ $\,=\,$ $\dfrac{p^2+1}{p}$

$\implies$ $\require{cancel} \sec{\theta}-\cancel{\tan{\theta}}$ $+$ $\require{cancel} \sec{\theta}+\cancel{\tan{\theta}}$ $\,=\,$ $\dfrac{p^2+1}{p}$

$\implies$ $2\sec{\theta}$ $\,=\,$ $\dfrac{p^2+1}{p}$

$\implies$ $\sec{\theta}$ $\,=\,$ $\dfrac{p^2+1}{2p}$

It can be used to find the value of cos function by using reciprocal identity of secant and cosine functions.

$\cos{\theta} \,=\, \dfrac{1}{\sec{\theta}}$

$\implies \cos{\theta} \,=\, \dfrac{1}{\dfrac{p^2+1}{2p}}$

$\implies \cos{\theta} \,=\, \dfrac{2p}{p^2+1}$

Evaluate sin function from cos function

The value of sin function can be calculated by the Pythagorean identity of sin and cos functions.

$\sin{\theta} \,=\, \sqrt{1-\cos^2{\theta}}$

Now, replace the cos function by its value and proceed further to get the value of sin function.

$\implies \sin{\theta} \,=\, \sqrt{1-{\Bigg(\dfrac{2p}{p^2+1}\Bigg)}^2}$

$\implies \sin{\theta} \,=\, \sqrt{1-\dfrac{4p^2}{{(p^2+1)}^2}}$

$\implies \sin{\theta} \,=\, \sqrt{ \dfrac{{(p^2+1)}^2-4p^2}{{(p^2+1)}^2}}$

$\implies \sin{\theta} \,=\, \dfrac{\sqrt{{(p^2+1)}^2-4p^2}}{p^2+1}$

Expand square of sum of two terms by using a plus b whole square formula.

$\implies \sin{\theta} \,=\, \dfrac{\sqrt{{(p^2)}^2+1^2+2p^2-4p^2}}{p^2+1}$

$\implies \sin{\theta} \,=\, \dfrac{\sqrt{{(p^2)}^2+1^2-2p^2}}{p^2+1}$

The expression under square root in numerator can be simplified by using a-b whole square identity.

$\implies \sin{\theta} \,=\, \dfrac{\sqrt{{(p^2-1)}^2}}{p^2+1}$

$\,\,\, \therefore \,\,\,\,\,\, \sin{\theta} \,=\, \dfrac{p^2-1}{p^2+1}$

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