# Find the value of tan 22°30’.

Question:

Find the value of tan 22°30’.

[Hint: Let $\theta=45^{\circ}$, use $\left.\tan \frac{\theta}{2}=\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}=\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}=\frac{\sin \theta}{1+\cos \theta}\right]$

Solution:

Let, θ = 45°

As we need to find: tan 22°30’ = tan (θ/2)

We know that,

sin θ = cos θ = 1/√2 (for θ = 45°)

Since,

$\tan \frac{\theta}{2}=\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}$

Multiplying $2 \cos \theta / 2$ in numerator and denominator, we get,

$\Rightarrow \tan \frac{\theta}{2}=\frac{2 \cos \frac{\theta}{2} \sin ^{\theta} \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}$

By applying formula of T-ratios of multiple angles-

$\sin 2 x=2 \sin x \cos x$

$\cos 2 x=2 \cos ^{2} x-1$ or $1+\cos 2 x=2 \cos ^{2} x$

$\therefore \tan \frac{\theta}{2}=\frac{\sin \theta}{1+\cos \theta}$

$\Rightarrow \tan 22^{\circ} 30^{\circ}=\frac{\sin 45^{\circ}}{1+\cos 45^{\circ}}$

$=\frac{\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}}$

$=\frac{1}{\sqrt{2}+1}$

By rationalizing the term, we get,

$\Rightarrow \tan 22^{\circ} 30^{\circ}=\frac{1}{\sqrt{2}+1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1}=\frac{\sqrt{2}-1}{(\sqrt{2})^{2}-1}$

Therefore, tan 22°30’ = √2 – 1