Question

please tell the answer faster ​

Answer 1

Answer:

169/49

Explanation:

=$\frac{2(1+SinA(1-SinA)}{2(1+CosA)(1-CosA)}$

=$\frac{(1-SinA)(1+SinA)}{(1+CosA)(1-CosA)}$

Since, [a-b][a+b]= a^2-b^2

=$\frac{1^{2}- Sin^{2} A }{1^{2}- Cos^{2}A }$

Since,$Sin^{2} A+Cos^{2} A=1$

=$\frac{Cos^{2} A}{Sin^{2} A}$

Since, Tan A= Sin A/ Cos A and Cot a= 1/ TanA

Cot A= Cos A/Tan A

Now,

=$\frac{(Cos A)( CosA)}{(SinA)(SinA)}$

=(CotA)(CotA)

=$Cot^{2}A$

Hence,

$\frac{2(1+SinA(1-SinA)}{2(1+CosA)(1-CosA)}$=$Cot^{2}A$

Given that Cot A= 13/7,

$Cot^{2} A= (\frac{13}{7})^{2}$

=$\frac{169}{49}$

P.S.: Please make do with theta as A because I could not find the symbol