Question

fimd the value of tan 22'30°

Answer 1

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Answer 2

Answer:

$\bf\tan22'30^{\circ}=\frac{1}{\sqrt{2}+1}$

Step-by-step explanation:

Let θ = 45°

$\implies\frac{\theta}{2}=22' 30^{\circ}\\\\Now,\tan\frac{\theta}{2}=\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}\\\\\text{Multiply both numerator and denominator by }2\cos\frac{\theta}{2}\\\\\implies \tan\frac{\theta}{2}=\frac{2cos\frac{\theta}{2}\times\sin\frac{\theta}{2}}{2cos\frac{\theta}{2}\times\cos\frac{\theta}{2}}\\\\\implies \tan\frac{\theta}{2}=\frac{\sin\theta}{2(\frac{1}{2}\times (1+\cos^2\frac{\theta}{2}))}\\\\\implies\tan\frac{\theta}{2}=\frac{\sin\theta}{2\cos\theta}\\\\\theta=45^{\circ}$

$\implies\tan\frac{\theta}{2}=\frac{\sin45^{\circ}}{1+\cos45^{\circ}} \\\\\bf\implies\tan22'30^{\circ}=\frac{1}{\sqrt{2}+1}$