Login/Sign Up
Home » If m sin θ = n sin (θ + 2α), then prove that tan (θ + α) cot α = (m + n)/(m – n) [Hints: Express sin(θ + 2α) / sinθ = m/n and apply componendo and dividend]
If m sin θ = n sin (θ + 2α), then prove that tan (θ + α) cot α = (m + n)/(m – n) [Hints: Express sin(θ + 2α) / sinθ = m/n and apply componendo and dividend]
CBSE | Class 11 | Maths | NCERT Exemplar | Trigonometric Functions
If m sin θ = n sin (θ + 2α), then prove that tan (θ + α) cot α = (m + n)/(m – n) [Hints: Express sin(θ + 2α) / sinθ = m/n and apply componendo and dividend]

Solution:

As per the question,

m \sin \theta=n \sin (\theta+2 a)

We need to prove:

\tan (\theta+\alpha) \cot \alpha=(m+n) /(m-n)

Proof:

\begin{array}{l} m \sin \theta=n \sin (\theta+2 a) \\ \Rightarrow \sin (\theta+2 \alpha) / \sin \theta=m / n \end{array}

Now, apply the componendo-dividendo rule, so we have,

\Rightarrow \frac{\sin (\theta+2 \alpha)+\sin \theta}{\sin (\theta+2 \alpha)-\sin \theta}=\frac{m+n}{m-n}

Using the transformation formula of T-ratios,

It is known that,

\sin A+\sin B=2 \sin ((A+B) / 2) \cos ((A-B) / 2)

And,

\sin A-\sin B=2 \cos ((A+B) / 2) \sin ((A-B) / 2)

By using the formula, we have,

\frac{2 \sin \left(\frac{2 \theta+2 \alpha}{2}\right) \cos \left(\frac{\theta+2 \alpha-\theta}{2}\right)}{2 \cos \left(\frac{2 \theta+2 \alpha}{2}\right) \sin \left(\frac{\theta+2 \alpha-\theta}{2}\right)}=\frac{\mathrm{m}+\mathrm{n}}{\mathrm{m}-\mathrm{n}}

\Rightarrow \frac{\sin (\theta+\alpha) \cos (\alpha)}{\cos (\theta+\alpha) \sin (\alpha)}=\frac{\mathrm{m}+\mathrm{n}}{\mathrm{m}-\mathrm{n}}
\{\because \tan \mathrm{x}=(\sin \mathrm{x}) /(\cos \mathrm{x})\}

\Rightarrow \tan (\theta+\alpha) \cot \alpha=\frac{\mathrm{m}+\mathrm{n}}{\mathrm{m}-\mathrm{n}}

So, \tan (\theta+\alpha) cot \alpha=(m+n) /(m-n)

As a result, hence proved.

i

More Material