A) True
B) False
C) Cannot say
D) Partially True/False
Correct Answer: B
Solution :
(False) We have, \[\frac{1-\sin \theta }{1+\sin \theta }=\frac{\left( 1-\sin \theta \right)\left( 1-\sin \theta \right)}{\left( 1+\sin \theta \right)\left( 1-\sin \theta \right)}\] \[\left[ \because \,\,rationalise \right]\] \[\Rightarrow \,\,\frac{{{\left( 1-\sin \theta \right)}^{2}}}{1-{{\sin }^{2}}\theta }={{\left( \frac{1-\sin \theta }{\cos \theta } \right)}^{2}}\] \[\left[ \because \,{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \right]\] \[={{\left( \sec \theta -\tan \theta \right)}^{2}}\] \[LHS\ne RHS\]
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