Question

prove sinA+ cos A/ sinA-cosA+ sinA-cosA/sinA+cosA=2/sin^2A-cos^2A

Answer 1

Answer:

$\bf\frac{sin\;A+cos\;A}{sin\;A-cos\;A}+\frac{sin\;A-cos\;A}{sin\;A+cos\;A}=\frac{2}{sin^2A-cos^2A}$

Step-by-step explanation:

Given:

$\frac{sin\;A+cos\;A}{sin\;A-cos\;A}+\frac{sin\;A-cos\;A}{sin\;A+cos\;A}$

$=\frac{(sin\;A+cos\;A)^2+(sin\;A-cos\;A)^2}{(sin\;A-cos\;A)(sin\;A+cos\;A)}$

Using the following algebraic identities

$\implies\boxed{\begin{minipage}{5cm} \bf(a+b)^2=a^2+b^2+2ab\\\\(a-b)^2=a^2+b^2-2ab\\\\(a-b)(a+b)=a^2-b^2 \end{minipage}}$

we get

$=\frac{sin^2A+cos^2A+2\,sin\,A\,cos\,A+sin^2A+cos^2A-2\,sin\,A\,cos\,A}{sin^2A-cos^2A}$

Using $\boxed{\bf\,sin^2\theta+cos^2\theta=1}$

$=\frac{1+2\,sin\,A\,cos\,A+1-2\,sin\,A\,cos\,A}{sin^2A-cos^2A}$

$=\frac{2}{sin^2A-cos^2A}$

$\implies\boxed{\bf\frac{sin\;A+cos\;A}{sin\;A-cos\;A}+\frac{sin\;A-cos\;A}{sin\;A+cos\;A}=\frac{2}{sin^2A-cos^2A}}$

Find more:

Prove that: Sin^3A+cos^3A= (sinA+cosA)(1-sinAcosA)

https://brainly.in/question/5667216

Answer 2

Answer:

Do see the image for the correct answer

Hope it helps!!!!!!!!!!!!!!!!!!!