Question

prove the image attached above...

please don't answer If you don't know ​

Answer 1

$\sqrt{\frac{secA-tanA}{secA+tanA} }$

Multiply numerator and denominator with $secA-tanA$

We get,

$\sqrt{\frac{(secA-tanA)^{2} }{(secA- tanA)(secA+tanA) } }$

Now using the formula $(a-b)(a+b)=a^{2}- b^{2}$

$\sqrt{\frac{(secA-tanA)^{2} }{sec^{2}A- tan^{2}A } }$

We know that ${sec^{2}A- tan^{2}A } }$

So, $\sqrt{\frac{(secA-tanA)^{2} }{sec^{2}A- tan^{2}A } }$ =$secA-tanA$

Hence, LHS=RHS