Question

Prove that sinA+cosA /sinA cos A+ sinA-cosA/ sinA+cosA=2/sin square-cos square=2/2 sin squareA-1=2/1-2cos squareA

Answer 1

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How to prove the identity `sin^2x + cos^2x = 1` ?

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EXPERT ANSWERS

PATTYSTEPHENS | CERTIFIED EDUCATOR

The equation of the unit circle is x^2+y^2=1.

All points on this circle have coordinates that make this equation true. 

For any random point (x, y) on the unit circle, the coordinates can be represented by (cos `theta` , sin `theta` ) where `theta` is the degrees of rotation from the positive x-axis (see attached image). 

By substituting cos `theta` = x and sin `theta` = y into the equation of the unit circle, we can see that (cos `theta` )^2 + (sin `theta` )^2 = 1. 

Answer 2

Here's your Answer

To Prove
$\tt\dfrac{sinA+cosA} {sinA-cosA}$ + $\tt\dfrac{sinA - cosA} {sinA+cosA}$ =$\tt\dfrac{2} {sin^2-cos^2}$ = $\tt\dfrac{2} {2sin^2-1}$=$\tt\dfrac{2} {1-2cos^2}$

Solution:

$\tt\dfrac{(sinA+cos) ^2 + (sinA-cosA) ^2} {sin^2A - cos^2}$

=>$\tt\dfrac{sin^2+cos^2+2sin×cos + sin^2+cos^2-2sin×cos} {sin^2 - cos^2}$

=>$\tt\dfrac{2sin^2 + 2cos^2} {sin^2 - cos^2}$

=>$\tt\dfrac{2(sin^2 + cos^2)} {sin^2 - cos^2}$

=>$\tt\dfrac{2} {sin^2-cos^2}$(1st Proved)

$\tt\dfrac{2} {sin^2-cos^2}$

=>$\tt\dfrac{2} {sin^2+sin^2-1}$

$\tt\dfrac{2} {2sin^2-1}$(2nd Proved)

$\tt\dfrac{2} {2sin^2-1}$

=>$\tt\dfrac{2} {(1-cos^2)+(-cos^2) }$

$\tt\dfrac{2} {1-2cos^2}$(3rd Proved)

$Hope\: it\: helps\: you\: :)$