Question

If $\tan \theta +\sin \theta =m\text{ and }\tan \theta -\sin \theta =n$ then prove that ${{m}^{2}}-{{n}^{2}}=4\sqrt{mn}$.

Answer 1

In this question, we are given the values of m and n in terms of trigonometric function. We need to prove ${{m}^{2}}-{{n}^{2}}=4\sqrt{mn}$. For this, we will solve the left side and the right side separately and then prove them equal. We will use the following properties.

& \left( i \right){{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\\ & \left( ii \right){{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\\ & \left( iii \right){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) \\\ & \left( iv \right)\tan \theta =\dfrac{\sin \theta }{\cos \theta } \\\ & \left( v \right){{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\\ \end{aligned}$$ **Complete step by step answer:** Here we are given the values of m and n in the form of trigonometric function as, $\tan \theta +\sin \theta =m\text{ and }\tan \theta -\sin \theta =n\cdots \cdots \left( 1 \right)$. We need to prove ${{m}^{2}}-{{n}^{2}}=4\sqrt{mn}$. For this let us first simplify the left side of the equation, we have the left side as ${{m}^{2}}-{{n}^{2}}$. Putting in the values from (1) we get, ${{\left( \tan \theta +\sin \theta \right)}^{2}}-{{\left( \tan \theta -\sin \theta \right)}^{2}}$. Now let us use expansion of algebraic function i.e. ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\text{ and }{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ we get, ${{\tan }^{2}}\theta +{{\sin }^{2}}\theta +2\tan \theta \sin \theta -\left( {{\tan }^{2}}\theta +{{\sin }^{2}}\theta -2\tan \theta \sin \theta \right)$. Simplifying we get, ${{\tan }^{2}}\theta +{{\sin }^{2}}\theta +2\tan \theta \sin \theta -{{\tan }^{2}}\theta -{{\sin }^{2}}\theta +2\tan \theta \sin \theta $. Cancelling the terms of ${{\tan }^{2}}\theta \text{ and }{{\sin }^{2}}\theta $ we get, $4\tan \theta \sin \theta $. Therefore the left side ${{m}^{2}}-{{n}^{2}}=4\tan \theta \sin \theta $. Now let us solve the right side of the given equation and prove it to be equal to $4\tan \theta \sin \theta $ so that the left side is equal to the right side. We have the right side as, $4\sqrt{mn}$. Putting in the values from (1) we get, $4\sqrt{\left( \tan \theta +\sin \theta \right)\left( \tan \theta -\sin \theta \right)}$. We know that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ using this we get, $4\sqrt{{{\tan }^{2}}\theta -{{\sin }^{2}}\theta }$. Now let us use the property $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ we get, $4\sqrt{\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }-{{\sin }^{2}}\theta }$. Taking ${{\sin }^{2}}\theta $ common inside the square root term we get, $4\sqrt{{{\sin }^{2}}\theta \left( \dfrac{1}{{{\cos }^{2}}\theta }-1 \right)}$. Taking LCM of ${{\cos }^{2}}\theta $ we get, $4\sqrt{{{\sin }^{2}}\theta \left( \dfrac{1-{{\cos }^{2}}\theta }{{{\cos }^{2}}\theta } \right)}$. Now we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ which can be written as ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $ so using this in the above equation we get, $4\sqrt{{{\sin }^{2}}\theta \cdot \dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }}$. Now again using $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ we get, $4\sqrt{{{\sin }^{2}}\theta {{\tan }^{2}}\theta }$. As we know that squares and square root cancel each other, so we get, $$4\sin \theta \tan \theta \Rightarrow 4\tan \theta \sin \theta $$. So the right side of the equation is also equal to $$4\tan \theta \sin \theta $$. Therefore, $${{m}^{2}}-{{n}^{2}}=4\tan \theta \sin \theta =4\sqrt{mn}$$ hence $${{m}^{2}}-{{n}^{2}}=4\sqrt{mn}$$. Hence proved. **Note:** Students should keep in mind all the trigonometric properties before solving this sum. Take care of the signs while applying algebraic properties. There are many ways to solve a sum so students can use different ways also to solve this sum. Take care of the signs while using trigonometric properties.