Question

# Prove the following questions : $$\dfrac{tan \,A}{ 1 + sec\, A} - \dfrac{tan \,A}{1 - sec\, A} = 2 \, cosec A$$

Solution

## We will start from LHS . [We can take $$tan \, A$$ as common and $$(1 - sec \, A0(1 + sec\,A)$$ makes $$(1 - sec^2A = tan^2 A)$$ so formulae will be used.] LHS $$= \dfrac{tan \,A}{ 1 + sec\, A} - \dfrac{tan \,A}{1 - sec\, A}$$       $$= tan \, A \left [ \dfrac{1}{(1 + sec\,A)} - \dfrac{1}{(1 - sec\,A)} \right ]$$  [Taking $$tan \, A$$ common]     $$= tan \, A \left [ \dfrac{1 - sec\,A - (1 + sec\,A)}{(1 + sec\,A )(1 - sec\,A)} \right ]$$  {Taking LCM)     $$= tan \,A \left [ \dfrac{-2 \, sec\,A}{(1 - sec^2A)} \right ] = \dfrac{tan \, A(-2\, sec\,A)}{-(sec^2A - 1)}$$     $$= \dfrac{-tan \, A \, 2\, sec\,A}{-tan^2A} = \dfrac{2 \, sec\,A}{tan\,A}$$     $$= \dfrac{2 \times \dfrac{1}{cos\,A}}{\dfrac{sin\,A}{cos\,A}}$$     $$= \dfrac{2}{sin\,A} = 2 \, cosec \,A =$$ RHS Hence , proved . Mathematics

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