Name: Trigonometric Ratios of Complementary Angles
Uploaded: 2019-07-19
Duration: 9 min 3 s
Description: What is the connection between the trigonometric ratios of an angle and its complementary angle? This video not only gives you that but helps you derive them...

Question

If

\tan θ = \frac{m}{n},

show that

\frac{m \sin θ - n \cos θ}{m \sin θ + n \cos θ} = \frac{m^{2} - n^{2}}{m^{2} + n^{2}} .

Accepted Answer

We have $\tan θ = \frac{m}{n}$

Let a right $△ A B C,$ in which $∠ A = θ °$ and $∠ B = 90 °$

$\tan θ = \frac{Perpendicular}{Base} = \frac{B C}{A B} = \frac{m}{n}$

Let $A B = (n) k,$ then $B C = (m) k,$ where $k$ is a positive number.

Using Pythagoras theorem,

$A C^{2} = A B^{2} + B C^{2} A C = \sqrt{A B^{2} + B C^{2}} A C = \sqrt{{(n k)}^{2} + {(m k)}^{2}} A C = \sqrt{n^{2} k^{2} + m^{2} k^{2}} = k \sqrt{m^{2} + n^{2}}$

Now, $\sin θ = \frac{Perpendicular}{Hypotenuse} = \frac{B C}{A C} = \frac{m k}{k \sqrt{m^{2} + n^{2}}} = \frac{m}{\sqrt{m^{2} + n^{2}}} . . . (i)$

and $\cos θ = \frac{Base}{Hypotenuse} = \frac{A B}{A C} = \frac{n k}{k \sqrt{m^{2} + n^{2}}} = \frac{n}{\sqrt{m^{2} + n^{2}}} . . . (i i)$

$∴ m \sin θ - n \cos θ = m \times \frac{m}{\sqrt{m^{2} + n^{2}}} - n \times \frac{n}{\sqrt{m^{2} + n^{2}}} = \frac{m^{2}}{\sqrt{m^{2} + n^{2}}} - \frac{n^{2}}{\sqrt{m^{2} + n^{2}}} = \frac{m^{2} - n^{2}}{\sqrt{m^{2} + n^{2}}} . . . (i i i)$

and $m \sin θ + n \cos θ = m \times \frac{m}{\sqrt{m^{2} + n^{2}}} + n \times \frac{n}{\sqrt{m^{2} + n^{2}}} = \frac{m^{2}}{\sqrt{m^{2} + n^{2}}} + \frac{n^{2}}{\sqrt{m^{2} + n^{2}}} = \frac{m^{2} + n^{2}}{\sqrt{m^{2} + n^{2}}} . . . (i v)$

So, $\frac{m \sin θ - n \cos θ}{m \sin θ + n \cos θ} = \frac{\frac{m^{2} - n^{2}}{\sqrt{m^{2} + n^{2}}}}{\frac{m^{2} + n^{2}}{\sqrt{m^{2} + n^{2}}}}$ [from equation $(i i i)$ and $(i v)$ ]

$= \frac{m^{2} - n^{2}}{\sqrt{m^{2} + n^{2}}} \times \frac{\sqrt{m^{2} + n^{2}}}{m^{2} + n^{2}} = \frac{m^{2} - n^{2}}{m^{2} + n^{2}}$

Thus, $\frac{m \sin θ - n \cos θ}{m \sin θ + n \cos θ} = \frac{m^{2} - n^{2}}{m^{2} + n^{2}} .$

If $\tan θ = \frac{m}{n},$ show that $\frac{m \sin θ - n \cos θ}{m \sin θ + n \cos θ} = \frac{m^{2} - n^{2}}{m^{2} + n^{2}} .$

Important Questions on Introduction To Trigonometry

Learn and Prepare for any exam you want !

If tan θ=mn , show that m sin θ-n cos θm sin θ+n cos θ=m2-n2m2+n2.

Important Questions on Introduction To Trigonometry

Learn and Prepare for any exam you want !

If $\tan θ = \frac{m}{n},$ show that $\frac{m \sin θ - n \cos θ}{m \sin θ + n \cos θ} = \frac{m^{2} - n^{2}}{m^{2} + n^{2}} .$