Question

If $\sec \theta + \tan \theta = p$, prove that $\dfrac{{{p^2} - 1}}{{{p^2} + 1}} = \sin \theta$

Answer 1

We use the trigonometric identity of ${\sec ^2}\theta - {\tan ^2}\theta = 1$and open this identity using another identity $(a + b)(a - b) = {a^2} - {b^2}$. Substitute the given value in the equation and write the remaining value in terms of p. Add the two equations formed in terms of secant and tangent. Calculate the value of cosine of angle and use the identity ${\sin ^2}\theta = 1 - {\cos ^2}\theta$ to calculate the value of sine of angle in terms of p.

Complete step-by-step solution:
We are given that $\left( {\sec \theta + \tan \theta } \right) = p$...................… (1)
Since we know that ${\sec ^2}\theta - {\tan ^2}\theta = 1$
Then we can open this identity using another identity $(a + b)(a - b) = {a^2} - {b^2}$
$\Rightarrow {\sec ^2}\theta - {\tan ^2}\theta = (\sec \theta - \tan \theta )(\sec \theta + \tan \theta )$
Substitute the value of $\sec \theta + \tan \theta = p$ in right hand side from equation (1)
$\Rightarrow {\sec ^2}\theta - {\tan ^2}\theta = p(\sec \theta - \tan \theta )$
Substitute the value of ${\sec ^2}\theta - {\tan ^2}\theta = 1$in left hand side of the equation
$\Rightarrow 1 = p(\sec \theta - \tan \theta )$
Divide both sides by p
$\Rightarrow \dfrac{1}{p} = (\sec \theta - \tan \theta )$...............… (2)
Add equations (1) and (2)
$\Rightarrow \left( {\sec \theta + \tan \theta } \right) + (\sec \theta - \tan \theta ) = p + \dfrac{1}{p}$
Cancel same terms with opposite signs from left hand side of the equation
$\Rightarrow 2\sec \theta = p + \dfrac{1}{p}$
Take LCM in right hand side of the equation
$\Rightarrow 2\sec \theta = \dfrac{{{p^2} + 1}}{p}$
Divide both sides by 2
$\Rightarrow \sec \theta = \dfrac{{{p^2} + 1}}{{2p}}$
Substitute the value of $\sec \theta = \dfrac{1}{{\cos \theta }}$in left hand side of the equation
$\Rightarrow \dfrac{1}{{\cos \theta }} = \dfrac{{{p^2} + 1}}{{2p}}$
Take reciprocal on both sides of the equation
$\Rightarrow \cos \theta = \dfrac{{2p}}{{1 + {p^2}}}$ … (3)
Now we know that ${\sin ^2}\theta = 1 - {\cos ^2}\theta$then we can take square root on both sides of the equation and write
$\Rightarrow \sqrt {{{\sin }^2}\theta } = \sqrt {1 - {{\cos }^2}\theta }$
Cancel square root by square power in left hand side of the equation
$\Rightarrow \sin \theta = \sqrt {1 - {{\cos }^2}\theta }$
Substitute value of cosine from equation (3) in right hand side of the equation
$\Rightarrow \sin \theta = \sqrt {1 - {{\left( {\dfrac{{2p}}{{1 + {p^2}}}} \right)}^2}}$
Solve the term in under root
$\Rightarrow \sin \theta = \sqrt {1 - \dfrac{{4{p^2}}}{{1 + {p^4} + 2{p^2}}}}$
Take LCM of terms in square root
$\Rightarrow \sin \theta = \sqrt {\dfrac{{1 + {p^4} + 2{p^2} - 4{p^2}}}{{1 + {p^4} + 2{p^2}}}}$
$\Rightarrow \sin \theta = \sqrt {\dfrac{{1 + {p^4} - 2{p^2}}}{{1 + {p^4} + 2{p^2}}}}$
We can pair the terms in numerator using the identity ${(a - b)^2} = {a^2} + {b^2} - 2ab$and denominator using the identity ${(a + b)^2} = {a^2} + {b^2} + 2ab$
$\Rightarrow \sin \theta = \sqrt {\dfrac{{{{({p^2} - 1)}^2}}}{{{{({p^2} + 1)}^2}}}}$
Cancel square root by square power in right hand side of the equation
$\Rightarrow \sin \theta = \dfrac{{{p^2} - 1}}{{{p^2} + 1}}$
$\therefore$The value of $\dfrac{{{p^2} - 1}}{{{p^2} + 1}} = \sin \theta$
Hence Proved

Note: Alternate method:
We directly substitute the value of p in the equation $\dfrac{{{p^2} - 1}}{{{p^2} + 1}} = \sin \theta$
$\Rightarrow \dfrac{{{{(\sec \theta + \tan \theta )}^2} - 1}}{{{{(\sec \theta + \tan \theta )}^2} + 1}} = \sin \theta$
Substitute the value of $\sec \theta = \dfrac{1}{{\cos \theta }};\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
$\Rightarrow \dfrac{{{{(\dfrac{1}{{\cos \theta }} + \dfrac{{\sin \theta }}{{\cos \theta }})}^2} - 1}}{{{{(\dfrac{1}{{\cos \theta }} + \dfrac{{\sin \theta }}{{\cos \theta }})}^2} + 1}} = \sin \theta$
Take LCM in the bracket
$\Rightarrow \dfrac{{{{(\dfrac{{1 + \sin \theta }}{{\cos \theta }})}^2} - 1}}{{{{(\dfrac{{1 + \sin \theta }}{{\cos \theta }})}^2} + 1}} = \sin \theta$
Square the terms in numerator and denominator
$\Rightarrow \dfrac{{\dfrac{{1 + {{\sin }^2}\theta + 2\sin \theta - {{\cos }^2}\theta }}{{{{\cos }^2}\theta }}}}{{\dfrac{{1 + {{\sin }^2}\theta + 2\sin \theta + {{\cos }^2}\theta }}{{{{\cos }^2}\theta }}}} = \sin \theta$
Write LHS of the equation in simpler form
$\Rightarrow \dfrac{{1 + {{\sin }^2}\theta + 2\sin \theta - {{\cos }^2}\theta }}{{{{\cos }^2}\theta }} \times \dfrac{{{{\cos }^2}\theta }}{{1 + {{\sin }^2}\theta + 2\sin \theta + {{\cos }^2}\theta }} = \sin \theta$
Cancel same terms from numerator and denominator in left hand side of the equation
$\Rightarrow \dfrac{{1 + {{\sin }^2}\theta + 2\sin \theta - {{\cos }^2}\theta }}{{1 + {{\sin }^2}\theta + 2\sin \theta + {{\cos }^2}\theta }} = \sin \theta$
Substitute ${\cos ^2}\theta = 1 - {\sin ^2}\theta$in both numerator and denominator in left hand side of the equation
$\Rightarrow \dfrac{{1 + {{\sin }^2}\theta + 2\sin \theta - (1 - {{\sin }^2}\theta )}}{{1 + {{\sin }^2}\theta + 2\sin \theta + (1 - {{\sin }^2}\theta )}} = \sin \theta$
$\Rightarrow \dfrac{{1 + {{\sin }^2}\theta + 2\sin \theta - 1 + {{\sin }^2}\theta }}{{1 + {{\sin }^2}\theta + 2\sin \theta + 1 - {{\sin }^2}\theta }} = \sin \theta$
Add same terms
$\Rightarrow \dfrac{{2{{\sin }^2}\theta + 2\sin \theta }}{{2 + 2\sin \theta }} = \sin \theta$
Take $2\sin \theta$common from numerator and 2 from denominator
$\Rightarrow \dfrac{{2\sin \theta (\sin \theta + 1)}}{{2(1 + \sin \theta )}} = \sin \theta$
Cancel same factors from numerator and denominator in left hand side of the equation
$\Rightarrow \sin \theta = \sin \theta$
LHS is equal to RHS
Hence proved