Question

$\frac{1-\sin \theta }{1+\sin \theta }=1+2{\tan }^2\theta -2\tan \theta \sec \theta$
Say true or false

• A. True
  
• B. False
• C. Ambiguous
• D. Data insufficient
  

Answer 1

The correct option is A True

$RHS$
$=1+2\frac{si{n}^2\theta }{co{s}^2\theta }-2\frac{sin\theta }{co{s}^2\theta }$
$=\frac{co{s}^2\theta +2si{n}^2\theta -2sin\theta }{co{s}^2\theta }$
$=\frac{1+si{n}^2\theta -2sin\theta }{1-si{n}^2\theta }$
$=\frac{(1-sin\theta {)}^2}{(1-sin\theta )(1+sin\theta )}$
$=\frac{1-sin\theta }{1+sin\theta }$
$=LHS$
Hence proved.