Question

Please HELP ME TO SOLVE THIS
sinA-2sin^3A/2cos^3A-cosA = tanA

Root1-sinA/1+sinA = cosA/1+sinA
PLEASE ANSWER CORRECTLY ​

Answer 1

Answer:

{sinA(1-2sin²A) }/{cosA(2cos²A-1)}=tanA(Taking sinA and Cos A as common...

--->tanA[1-2(1-cos²A)/(2cos²A-1)]=tanA

==tanA[1-2+2cos²A/2cos²A-1]=tanA

===tan A[2cos²A-1/2cos²A-1]=tanA

== therefore---TanA=rhs(TanA)

== hence proved

√1-sinA/√1+sinA===By multiplying both by √1+sinA...

√1-sin²A/√(1+sinA)²===√cos²A/√(1+sin²A)²===

Now open the Roots

==>>cosA/1+sinA=Rhs(cosA/1+sinA)

Answer 2

Question 1

$\frac{sina - 2 {sin}^{3} a}{2 {cos}^{3} a - cosa} \\ \\ = \frac{sina(1 - 2 {sin}^{2} a)}{cosa(2 {cos}^{2}a - 1) } \\ \\ = \frac{sina(cos2a)}{cosa(cos2a)} \\ \\ = \frac{sina}{cosa} \\ \\ = tana$

Question 2

$\sqrt{ \frac{1 - sina}{1 + sina} } \\ \\ = \sqrt{ \frac{1 - sina}{1 + sina} \times \frac{1 + sina}{1 + sina} } \\ \\ = \sqrt{ \frac{ {cos}^{2}a }{( {1 + sina})^{2} } } \\ \\ = \frac{cosa}{1 + sina}$