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Question: If \(\sec a+\tan a=b\), then find \(\sin a\)....

If sec⁡a+tan⁡a=b\sec a+\tan a=b, then find sin⁡a\sin a.

Explanation

Solution

We will write the secant and tangent functions in terms of sine and cosine function. Then we will use the identity sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1 and replace the cosine function in terms of sine function. Then we will obtain a quadratic equation in terms of the sine function. Comparing it with a standard quadratic equation, we will find the value of sin⁡a\sin a by using the quadratic formula.

Complete step by step answer:
We know that sec⁡a=1cos⁡a\sec a=\dfrac{1}{\cos a} and tan⁡a=sin⁡acos⁡a\tan a=\dfrac{\sin a}{\cos a}. Substituting these values in the given equation, we get
1cos⁡a+sin⁡acos⁡a=b\dfrac{1}{\cos a}+\dfrac{\sin a}{\cos a}=b
Simplifying the above equation, we get
1+sin⁡acos⁡a=b\dfrac{1+\sin a}{\cos a}=b
Substituting cos⁡a=1−sin⁡2a\cos a=\sqrt{1-{{\sin }^{2}}a} in the above equation, we have
1+sin⁡a1−sin⁡2a=b ∴1+sin⁡a=b1−sin⁡2a \begin{aligned} & \dfrac{1+\sin a}{\sqrt{1-{{\sin }^{2}}a}}=b \\\ & \therefore 1+\sin a=b\sqrt{1-{{\sin }^{2}}a} \\\ \end{aligned}
Now, squaring both sides of the above equation, we get
(1+sin⁡a)2=(b1−sin⁡2a)2{{\left( 1+\sin a \right)}^{2}}={{\left( b\sqrt{1-{{\sin }^{2}}a} \right)}^{2}}
On the left hand side, we will expand the square of the bracket using the identity (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab in the following manner,
1+sin⁡2a+2sin⁡a=b2(1−sin⁡2a) ⇒1+sin⁡2a+2sin⁡a=b2−b2sin⁡2a ⇒1+sin⁡2a+2sin⁡a−b2+b2sin⁡2a=0 ⇒(1+b2)sin⁡2a+2sin⁡a+(1−b2)=0 \begin{aligned} & 1+{{\sin }^{2}}a+2\sin a={{b}^{2}}\left( 1-{{\sin }^{2}}a \right) \\\ & \Rightarrow 1+{{\sin }^{2}}a+2\sin a={{b}^{2}}-{{b}^{2}}{{\sin }^{2}}a \\\ & \Rightarrow 1+{{\sin }^{2}}a+2\sin a-{{b}^{2}}+{{b}^{2}}{{\sin }^{2}}a=0 \\\ & \Rightarrow \left( 1+{{b}^{2}} \right){{\sin }^{2}}a+2\sin a+\left( 1-{{b}^{2}} \right)=0 \\\ \end{aligned}
We have obtained a quadratic equation in terms of sin⁡a\sin a. Comparing this with the standard quadratic equation ax2+bx+c=0a{{x}^{2}}+bx+c=0, we have a=1+b2a=1+{{b}^{2}}, b=2b=2and c=1−b2c=1-{{b}^{2}}. Now, we will use the quadratic formula which is x=−b±b2−4ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} to find the value of sin⁡a\sin a in the following manner,
sin⁡a=−2±(2)2−4(1+b2)(1−b2)2(1+b2)\sin a=\dfrac{-2\pm \sqrt{{{\left( 2 \right)}^{2}}-4\left( 1+{{b}^{2}} \right)\left( 1-{{b}^{2}} \right)}}{2\left( 1+{{b}^{2}} \right)}
We can write (1+b2)(1−b2)=1−b4\left( 1+{{b}^{2}} \right)\left( 1-{{b}^{2}} \right)=1-{{b}^{4}} since we know that (a+b)(a−b)=a2−b2\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}.
So, we have the following,
sin⁡a=−2±4−4(1−b4)2(1+b2)\sin a=\dfrac{-2\pm \sqrt{4-4\left( 1-{{b}^{4}} \right)}}{2\left( 1+{{b}^{2}} \right)}
Simplifying the above expression, we get
sin⁡a=−2±4(1−1+b4)2(1+b2) ⇒sin⁡a=−2±2b42(1+b2) ⇒sin⁡a=−1±b21+b2 \begin{aligned} & \sin a=\dfrac{-2\pm \sqrt{4\left( 1-1+{{b}^{4}} \right)}}{2\left( 1+{{b}^{2}} \right)} \\\ & \Rightarrow \sin a=\dfrac{-2\pm 2\sqrt{{{b}^{4}}}}{2\left( 1+{{b}^{2}} \right)} \\\ & \Rightarrow \sin a=\dfrac{-1\pm {{b}^{2}}}{1+{{b}^{2}}} \\\ \end{aligned}

So, we have the value of sin⁡a=−1+b21+b2\sin a=\dfrac{-1+{{b}^{2}}}{1+{{b}^{2}}}and sin⁡a=−1−b21+b2=−(1+b2)1+b2=−1\sin a=\dfrac{-1-{{b}^{2}}}{1+{{b}^{2}}}=\dfrac{-\left( 1+{{b}^{2}} \right)}{1+{{b}^{2}}}=-1.

Note: It is useful to know the trigonometric identities and the algebraic identities for this type of question. The calculations or derivations are lengthy so it is beneficial if we do the calculations explicitly. This way, we can avoid making any minor mistakes and obtain the correct answer. The key aspect in this was to identify the equation obtained in terms of the sine function as a quadratic equation.