Question

If we are given two statements as $\cot \theta \left( 1+\sin \theta \right)=4m$ and $\cot \theta \left( 1-\sin \theta \right)=4n$, then using the trigonometric identities and properties, prove that ${{\left( {{m}^{2}}-{{n}^{2}} \right)}^{2}}=mn$,

Answer 1

Hint: We can start solving this by considering the RHS, i.e. $m\times n$ first. We have the values of $4m$ and $4n$ in the question from which we can get the values of m and n. Then after substitution, we will get a result. Then, we can solve the LHS, that is ${{\left( {{m}^{2}}-{{n}^{2}} \right)}^{2}}$ and then find its result.

Complete step-by-step answer:

It is given in the question that $\cot \theta \left( 1+\sin \theta \right)=4m$ and $\cot \theta \left( 1-\sin \theta \right)=4n$ then, $m=\dfrac{\cot \theta \left( 1+\sin \theta \right)}{4}$ and $n=\dfrac{\cot \theta \left( 1-\sin \theta \right)}{4}$. Using these values we have to prove that ${{\left( {{m}^{2}}-{{n}^{2}} \right)}^{2}}=mn$. Firstly, we will simplify RHS.

In RHS we have $m\times n$, we know that $m=\dfrac{\cot \theta \left( 1+\sin \theta \right)}{4}$ and $n=\dfrac{\cot \theta \left( 1-\sin \theta \right)}{4}$, therefore $m\times n=\dfrac{\cot \theta \left( 1+\sin \theta \right)}{4}\times \dfrac{\cot \theta \left( 1-\sin \theta \right)}{4}$ = $\dfrac{\cot \theta +\cot \theta \sin \theta }{4}\times \dfrac{\cot \theta -\cot \theta \sin \theta }{4}$, simplifying further by using the generic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ we get $m\times n=\dfrac{{{\cot }^{2}}\theta -{{\cot }^{2}}\theta {{\sin }^{2}}\theta }{16}$.

Now, we know that $\left( 1-{{\sin }^{2}}\theta \right)={{\cos }^{2}}\theta $, we get $m\times n=\dfrac{{{\cot }^{2}}\theta \times {{\cos }^{2}}\theta }{16}$, on further simplifying we get $mn=\dfrac{{{\cos }^{2}}\theta {{\cos }^{2}}\theta }{16{{\sin }^{2}}\theta }$ or finally $mn=\dfrac{{{\cos }^{4}}\theta }{16{{\sin }^{2}}\theta }....(i)$.

Now, we will solve RHS, we get - ${{\left( {{m}^{2}}-{{n}^{2}} \right)}^{2}}={{\left[ \left( \dfrac{{{\cot }^{2}}\theta {{\left( 1+\sin \theta \right)}^{2}}}{{{\left( 4 \right)}^{2}}} \right)-\left( \dfrac{{{\cot }^{2}}\theta {{\left( 1-\sin \theta \right)}^{2}}}{{{\left( 4 \right)}^{2}}} \right) \right]}^{2}}$on simplifying further, we get the following expression \[=\dfrac{{{\left( {{\cot }^{2}}\theta \right)}^{2}}}{{{\left( 16 \right)}^{2}}}\left[ {{\left( 1+\sin \theta \right)}^{2}}-{{\left( 1-\sin \theta \right)}^{2}} \right]\], solving the bracket further by opening the square using the generic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and the identity ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$, we get ${{\left( {{m}^{2}}-{{n}^{2}} \right)}^{2}}=\dfrac{{{\cot }^{4}}\theta }{{{\left( 16 \right)}^{2}}}{{\left[ \left( 1+{{\sin }^{2}}\theta +2\sin \theta \right)-\left( 1+{{\sin }^{2}}\theta -2\sin \theta \right) \right]}^{2}}$ therefore cancelling the common terms from the expression, we get reduced form as ${{\left( {{m}^{2}}-{{n}^{2}} \right)}^{2}}=\dfrac{{{\cot }^{4}}\theta }{{{\left( 16 \right)}^{2}}}{{\left[ 4\sin \theta \right]}^{2}}$.

Finally, reducing the above expression using $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ we get into the last step as follows \[{{\left( {{m}^{2}}-{{n}^{2}} \right)}^{2}}=\dfrac{{{\cos }^{4}}\theta }{{{\left( 16 \right)}^{2}}{{\sin }^{4}}\theta }16{{\sin }^{2}}\theta =\dfrac{{{\cos }^{4}}\theta }{16{{\sin }^{2}}\theta }...(ii)\] . Therefore from equation 1 and equation 2 we can see that both LHS and RHS are equal or result in the same expression on some modifications. Hence \[LHS\text{ }=\text{ }RHS\], thus proved.


Note: This is a very basic question of trigonometry but usually students get stuck in multiplying the exponential powers. Majority of the students skip these types of questions in exams by just seeing the power of trigonometry functions, but it is not true if we go step by step to solve this we can solve it easily.