Question

If $\tan \theta + \sin \theta = m$ and $\tan \theta - \sin \theta = n$, then:
A. ${m^2} - {n^2} = 4mn$
B. ${m^2} - {n^2} = 4mn$
C. ${m^2} - {n^2} = {m^2} + {n^2}$
D. ${m^2} - {n^2} = 4\sqrt {mn} $

Answer 1

Hint:In the given problem, we are given two equations involving the trigonometric functions. The question requires thorough knowledge of trigonometric functions, formulae and identities. We solve the two equations using the method of elimination to find the values of trigonometric functions. Then, we use some trigonometric equations and formulae to get to the result and match the options.

Complete step by step answer:
In the given question, we are provided with the equations: $\tan \theta + \sin \theta = m$ and $\tan \theta - \sin \theta = n$. So, we have to find the values of both the trigonometric functions: tangent and sine using the two equations. So, we add up two equations. Hence, we get,
$ \Rightarrow \tan \theta + \sin \theta + \tan \theta - \sin \theta = m + n$
Now, adding up the like terms and simplifying the equation, we get,
$ \Rightarrow 2\tan \theta = m + n$
Dividing both the sides of the equation by $2$, we get,
$ \Rightarrow \tan \theta = \left( {\dfrac{{m + n}}{2}} \right)$
So, we get the value of tangent as $\left( {\dfrac{{m + n}}{2}} \right)$. Now, we find the value of sine by substituting the value of tangent in any of the two equations.
So, we get, $\tan \theta + \sin \theta = m$
$ \Rightarrow \left( {\dfrac{{m + n}}{2}} \right) + \sin \theta = m$
Taking all the terms except the sine function to right side of the equation, we get,
$ \Rightarrow \sin \theta = m - \left( {\dfrac{{m + n}}{2}} \right)$
Taking LCM of the denominators,
$ \Rightarrow \sin \theta = \left( {\dfrac{{2m - \left( {m + n} \right)}}{2}} \right)$
$ \Rightarrow \sin \theta = \left( {\dfrac{{m - n}}{2}} \right)$
So, we get the value of sine as $\left( {\dfrac{{m - n}}{2}} \right)$.

Now, we find the relations between the variables m and n by using a trigonometric identity or formula. So, we know the trigonometric identity \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]. So, we can find a relation between m and n by substituting the values of cosine and sine in the identity. Now, we will first have to find the value of $\cos \theta $. We know the trigonometric formula of tangent $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$. So, we substitute the values of tangent and sine in the formula to get the value of cosine. So, we get,
$ \Rightarrow \left( {\dfrac{{m + n}}{2}} \right) = \dfrac{{\left( {\dfrac{{m - n}}{2}} \right)}}{{\cos \theta }}$
Cross multiplying the terms of the equation, we get,
$ \Rightarrow \cos \theta = \dfrac{{\left( {\dfrac{{m - n}}{2}} \right)}}{{\left( {\dfrac{{m + n}}{2}} \right)}}$
Simplifying the value of cosine, we get,
$ \Rightarrow \cos \theta = \dfrac{{m - n}}{{m + n}}$
So, we get the value of cosine as $\left( {\dfrac{{m - n}}{{m + n}}} \right)$.

Now, we substitute the values of cosine and sine in the trigonometric equation \[{\sin ^2}\theta + {\cos ^2}\theta = 1\] to get the required relation between m and n, Hence, we get,
\[ \Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = 1\]
\[ \Rightarrow {\left( {\dfrac{{m - n}}{2}} \right)^2} + {\left( {\dfrac{{m - n}}{{m + n}}} \right)^2} = 1\]
Computing the squares of the terms, we get,
\[ \Rightarrow \left( {\dfrac{{{m^2} - 2mn + {n^2}}}{4}} \right) + \left( {\dfrac{{{m^2} - 2mn + {n^2}}}{{{m^2} + 2mn + {n^2}}}} \right) = 1\]
Taking the LCM of the denominators, we get,
\[ \Rightarrow \left( {\dfrac{{\left( {{m^2} - 2mn + {n^2}} \right)\left( {{m^2} + 2mn + {n^2}} \right) + 4\left( {{m^2} - 2mn + {n^2}} \right)}}{{4\left( {{m^2} + 2mn + {n^2}} \right)}}} \right) = 1\]
Expanding the numerator, we get,
\[ \Rightarrow \left( {\dfrac{{{m^4} - 2{m^3}n + {n^2}{m^2} + 2{m^3}n - 4{n^2}{m^2} + 2m{n^3} + {n^2}{m^2} - 2m{n^3} + {n^4} + 4{m^2} - 8mn + 4{n^2}}}{{4\left( {{m^2} + 2mn + {n^2}} \right)}}} \right) = 1\]

Cancelling the like terms with opposite signs, we get,
\[ \Rightarrow \left( {\dfrac{{{m^4} - 2{n^2}{m^2} + {n^4} + 4{m^2} - 8mn + 4{n^2}}}{{4\left( {{m^2} + 2mn + {n^2}} \right)}}} \right) = 1 \\ \]
Cross multiplying the terms of the equation,
\[ \Rightarrow {m^4} - 2{n^2}{m^2} + {n^4} + 4{m^2} - 8mn + 4{n^2} = 4{m^2} + 8mn + 4{n^2} \\ \]
\[ \Rightarrow {m^4} - 2{n^2}{m^2} + {n^4} - 8mn = 8mn \\ \]
Adding $8mn$ to both sides of the equation, we get,
\[ \Rightarrow {m^4} - 2{n^2}{m^2} + {n^4} = 16mn \\ \]
\[ \Rightarrow {\left( {{m^2}} \right)^2} - 2\left( {{n^2}} \right)\left( {{m^2}} \right) + {\left( {{n^2}} \right)^2} = 16mn \\ \]
Using the algebraic identity ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$, we get,
\[ \Rightarrow {\left( {{m^2} - {n^2}} \right)^2} = 16mn \\ \]
Taking square root on both sides of the equation,
\[ \Rightarrow \left( {{m^2} - {n^2}} \right) = \sqrt {16mn} \\ \]
\[ \therefore \left( {{m^2} - {n^2}} \right) = 4\sqrt {mn} \]
So, we get the relation between m and n as \[\left( {{m^2} - {n^2}} \right) = 4\sqrt {mn} \].

Hence, the correct answer is option D.

Note:There are six trigonometric ratios: $\sin \theta $, $\cos \theta $, $\tan \theta $, $\cos ec\theta $, $\sec \theta $and $\cot \theta $. Basic trigonometric identities include ${\sin ^2}\theta + {\cos ^2}\theta = 1$, ${\sec ^2}\theta = {\tan ^2}\theta + 1$ and $\cos e{c^2}\theta = {\cot ^2}\theta + 1$. These identities are of vital importance for solving any question involving trigonometric functions and identities. All the trigonometric ratios can be converted into each other using the simple trigonometric identities listed above. The given problem involves the use of trigonometric formulae and identities. Such questions require thorough knowledge of trigonometric conversions and ratios. Algebraic operations and rules like transposition rules come into significant use while solving such problems.