Is 2sec(2x)=tan(pi/4+x)+tan(pi/4-x) ?

The answer to whether 2sec(2x)=tan(pi/4+x)+tan(pi/4-x) is True

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prove 2sec(2x)=tan(pi/4+x)+tan(pi/4-x)

prove

2 sec (2 x) = tan (\frac{π}{4} + x) + tan (\frac{π}{4} - x)

True

Solution steps

2 sec (2 x) = tan (\frac{π}{4} + x) + tan (\frac{π}{4} - x)

Manipulating right side

tan (\frac{π}{4} + x) + tan (\frac{π}{4} - x)

Rewrite using trig identities

= \frac{cos ( x ) + sin ( x )}{cos ( x ) - sin ( x )} + tan (\frac{π}{4} - x)

Rewrite using trig identities

= \frac{cos ( x ) + sin ( x )}{cos ( x ) - sin ( x )} + \frac{cos ( x ) - sin ( x )}{cos ( x ) + sin ( x )}

Simplify

\frac{cos ( x ) + sin ( x )}{cos ( x ) - sin ( x )} + \frac{cos ( x ) - sin ( x )}{cos ( x ) + sin ( x )} : \frac{2 cos ^{2} ( x ) + 2 sin ^{2} ( x )}{( cos ( x ) - sin ( x ) ) ( cos ( x ) + sin ( x ) )}

= \frac{2 cos ^{2} ( x ) + 2 sin ^{2} ( x )}{( cos ( x ) - sin ( x ) ) ( cos ( x ) + sin ( x ) )}

Factor

\frac{2 cos ^{2} ( x ) + 2 sin ^{2} ( x )}{( cos ( x ) + sin ( x ) ) ( cos ( x ) - sin ( x ) )} : \frac{2 ( cos ^{2} ( x ) + sin ^{2} ( x ) )}{( cos ( x ) + sin ( x ) ) ( cos ( x ) - sin ( x ) )}

= \frac{( cos ^{2} ( x ) + sin ^{2} ( x ) ) \cdot 2}{( cos ( x ) + sin ( x ) ) ( cos ( x ) - sin ( x ) )}

Rewrite using trig identities

= \frac{2}{cos ( 2 x )}

Rewrite using trig identities

2 sec (2 x)

We showed that the two sides could take the same form

\Rightarrow True

sin (x) - csc (x) = - (cos (x)) (cot (x))

tan (θ) cot (θ) = cos^{2} (θ) + sin^{2} (θ)

cos (θ) = sin (3 θ + 62)

sin (π \cdot x) + sin (π \cdot (10 - x)) = 0

cot (2 θ) = \frac{1}{2} (tan (θ) - cot (θ))

Is 2sec(2x)=tan(pi/4+x)+tan(pi/4-x) ?
The answer to whether 2sec(2x)=tan(pi/4+x)+tan(pi/4-x) is True