Is 6cos^2(A)-5=1-6sin^2(A) ?

The answer to whether 6cos^2(A)-5=1-6sin^2(A) is True

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prove 6cos^2(A)-5=1-6sin^2(A)

prove

6 cos^{2} (A) - 5 = 1 - 6 sin^{2} (A)

True

Solution steps

6 cos^{2} (A) - 5 = 1 - 6 sin^{2} (A)

Manipulating left side

6 cos^{2} (A) - 5

Rewrite using trig identities

6 cos^{2} (A) - 5

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= 6 (1 - sin^{2} (A)) - 5

Simplify

6 (1 - sin^{2} (A)) - 5 : - 6 sin^{2} (A) + 1

6 (1 - sin^{2} (A)) - 5

Expand

6 (1 - sin^{2} (A)) : 6 - 6 sin^{2} (A)

6 (1 - sin^{2} (A))

Apply the distributive law:

a (b - c) = ab - a c

a = 6, b = 1, c = sin^{2} (A)

= 6 \cdot 1 - 6 sin^{2} (A)

Multiply the numbers:

6 \cdot 1 = 6

= 6 - 6 sin^{2} (A)

= 6 - 6 sin^{2} (A) - 5

Simplify

6 - 6 sin^{2} (A) - 5 : - 6 sin^{2} (A) + 1

6 - 6 sin^{2} (A) - 5

Group like terms

= - 6 sin^{2} (A) + 6 - 5

Add/Subtract the numbers:

6 - 5 = 1

= - 6 sin^{2} (A) + 1

= - 6 sin^{2} (A) + 1

= - 6 sin^{2} (A) + 1

= - 6 sin^{2} (A) + 1

= 1 - 6 sin^{2} (A)

We showed that the two sides could take the same form

\Rightarrow True

sin (x) + \frac{cot ( x )}{sec ( x )} = csc (x)

\frac{sin ( 2 θ )}{sin ( θ )} = 2

cos^{2} (x) - sin^{2} (x) + 1 = 2 cos^{2} (x)

- 2 sin (2 x + π) = - 2 sin (2 x - π)

cos (θ) (\frac{1}{1 - sin ( θ )}) = sec (θ) + tan (θ)

Is 6cos^2(A)-5=1-6sin^2(A) ?
The answer to whether 6cos^2(A)-5=1-6sin^2(A) is True