Prove that 1+ - 1+ + =1- . - Vidyadip

Question

Prove that $\frac{1+\sec\theta-\tan\theta}{1+\sec\theta+\tan\theta}=\frac{1-\sin\theta}{\cos\theta}.$

✓

Answer

Recall identity $\sec^2\theta-\tan^2\theta=1$
LHS $=\frac{1+\sec\theta-\tan\theta}{1+\sec\theta+\tan\theta}$
$=\frac{\sec^2\theta-\tan^2\theta+\sec\theta-\tan\theta}{1+\sec\theta+\tan\theta}$
$=\frac{(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)+(\sec\theta-\tan\theta)}{1+\sec\theta+\tan\theta}$
$[\because\text{a}^2-\text{b}^2=(\text{a}-\text{b})(\text{a}+\text{b})]$
$=\frac{(\sec\theta-\tan\theta)[\sec\theta+\tan\theta+1]}{(\sec\theta+\tan\theta+1)}$
$=\sec\theta-\tan\theta$
$=\frac{1}{\cos\theta}-\frac{\sin\theta}{\cos\theta}$
$=\frac{1-\sin\theta}{\cos\theta}=$ RHS
Hence, proved.

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