Question

If the following assumption is assumed $\tan \theta =\dfrac{a}{b}$, Using the trigonometry properties and formulas, show that $\dfrac{a\sin \theta -b\cos \theta }{a\sin \theta +b\cos \theta }=\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}$.

Answer 1

Hint: In a right angled triangle ABC, we have ${{\left( base \right)}^{2}}+{{\left( height \right)}^{2}}={{\left( hypotenuse \right)}^{2}}$ that is as shown in the following figure

Complete step-by-step answer:

Also, we know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and from the figure above we get $\sin \theta =\dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$ and also $\cos \theta =\dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$. Using all three relations we will try to show that LHS = RHS.

It is given in the question that $\tan \theta =\dfrac{a}{b}$ and we have to show that $\dfrac{a\sin \theta -b\cos \theta }{a\sin \theta +b\cos \theta }=\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}$. We know that in a right angled triangle, pythagoras theorem state that ${{\left( base \right)}^{2}}+{{\left( height \right)}^{2}}={{\left( hypotenuse \right)}^{2}}$.

Also we know that \[\sin \theta =\dfrac{height}{hypotenous}\], here we have \[\sin \theta =\dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\] and \[\cos \theta =\dfrac{base}{hypotenous}\], here we have, \[\cos \theta =\dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\]. We know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ = $\dfrac{\dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}}}{\dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}}}}$ , therefore we get $\tan \theta =\dfrac{a}{b}$.

Now, in LHS we are given the equation as $\dfrac{a\sin \theta -b\cos \theta }{a\sin \theta +b\cos \theta }$ Putting the value of $\sin \theta $ as $\dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$ and $\cos \theta $ as $\dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$ in LHS, we get $\dfrac{a\left( \dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right)-b\left( \dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right)}{a\left( \dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right)+b\left( \dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right)}$ = $\dfrac{\left( \dfrac{{{a}^{2}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right)-\left( \dfrac{{{b}^{2}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right)}{\left( \dfrac{{{a}^{2}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right)+\left( \dfrac{{{b}^{2}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right)}$ = $\dfrac{\dfrac{{{a}^{2}}-{{b}^{2}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}}}{\dfrac{{{a}^{2}}+{{b}^{2}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}}}$ cancelling $\sqrt{{{a}^{2}}+{{b}^{2}}}$ from numerator and denominator, we get $\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}$ also in our RHS we are given with expression as $\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}$.
Thus LHS = RHS, hence proved.

Note: This question can be solved in just few step if we go through this process – given that $\tan \theta =\dfrac{a}{b}$ also we know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ therefore we get $\sin \theta =a$ and $\cos \theta =b$ therefore from LHS of the given equation, on putting the above values we get $\dfrac{a\sin \theta -b\cos \theta }{a\sin \theta +b\cos \theta }=\dfrac{a\left( a \right)-b\left( b \right)}{a\left( a \right)+b\left( b \right)}=\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}=RHS$ which is equal to RHS of the given equation that is to be proved.
Hence proved.