Question

tan A -Sin A /sin2A
${sin}^{2}$
=
1+ CosA​

Answer 1

Step-by-step explanation:

Given:-

(Tan A - Sin A)/ Sin^2 A

To show:-

(Tan A - Sin A)/ Sin^2 A = Tan A/(1+Cos A)

Solution:-

LHS:-

(Tan A - Sin A)/ Sin^2 A

=>[(SinA/Cos A)-Sin A]/ Sin^2 A

=>Sin A[(1/Cos A)-1)]/ Sin^2 A

=>Sin A [(1-Cos A)/Cos A] / Sin^2 A

On cancelling Sin A then

=>[ (1-Cos A)/Cos A] / Sin A

=>(1-Cos A)/( Sin A Cos A)

On multiplying both numerator and denominator by (1+CosA) then

=>(1-Cos A) (1+Cos A)/[Sin A Cos A (1+Cos A)]

=>(1-Cos^2 A)/[Sin A Cos A (1+Cos A)]

Since (a+b)(a-b)=a^2-b^2

We know that

Sin^2 A + Cos^2 A = 1

=> Sin^2 A/[Sin A Cos A (1+Cos A)]

On cancelling SinA

=>SinA/[ Cos A (1+Cos A)]

=>(Sin A/Cos A)/(1+Cos A)

=>Tan A/(1+Cos A)

=>RHS

LHS = RHS

Hence, Proved

Answer:-

(Tan A - Sin A)/ Sin^2 A = Tan A/(1+Cos A)

Used formulae:-

• (a+b)(a-b)=a^2-b^2

• Sin^2 A + Cos^2 A = 1

• Tan A = Sin A/ Cos A