Math, asked by anuragpaul2204, 5 months ago

Prove that 1/ secA - tanA - 1/ cosA = 1/ cosA - 1/ secA + tanA​

Answers

Answered by anshika3834
1
Hope this helps.......
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Answered by sandy1816
0

LHS

 \frac{1}{secA - tanA}  -  \frac{1}{cosA}  \\  \\  =  \frac{1}{secA - tanA}  \times  \frac{secA + tanA}{secA + tanA}  -  \frac{1}{cosA}  \\  \\  =  \frac{secA + tanA}{ {sec}^{2}A -  {tan}^{2}  A}  - secA \\  \\  = secA + tanaA- secA \\  \\  = secA - (secA - tanA) \\  \\  = secA - ( \frac{secA - tanA}{ {sec}^{2}A -  {tan}^{2} A }  \\  \\  = secA -  \frac{1}{secA + tanA}  \\  \\  =  \frac{1}{cosA}  -  \frac{1}{secA + tanA}

=RHS.

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