tanθ + tan (90° – θ) = sec θ

Question:

tanθ + tan (90° – θ) = sec θ sec (90° – θ)

Solution:

LHS $=\tan \theta+\tan \left(90^{\circ}-\theta\right)$ $\left[\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right]$

$=\tan \theta+\cot \theta=\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}$

$=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta}$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right.$ and $\left.\cot \theta=\frac{\cos \theta}{\sin \theta}\right]$

$=\frac{1}{\sin \theta \cos \theta}$ $\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$

$=\sec \theta \operatorname{cosec} \theta$ $\left[\because \sec \theta=\frac{1}{\cos \theta}\right.$ and $\left.\cos \theta=\frac{1}{\sin \theta}\right]$

$=\sec \theta \sec \left(90^{\circ}-\theta\right)=\mathrm{RHS}$ $\left[\because \sec \left(90^{\circ}-\theta\right)=\operatorname{cosec} \theta\right]$

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