Is sin^2(90-b)+cos^2(b-450)=1 ?

The answer to whether sin^2(90-b)+cos^2(b-450)=1 is True

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prove sin^2(90-b)+cos^2(b-450)=1

prove

sin^{2} (9 0^{\circ} - b) + cos^{2} (b - 45 0^{\circ}) = 1

True

Solution steps

sin^{2} (9 0^{\circ} - b) + cos^{2} (b - 45 0^{\circ}) = 1

Manipulating left side

sin^{2} (9 0^{\circ} - b) + cos^{2} (b - 45 0^{\circ})

Rewrite using trig identities

= sin^{2} (9 0^{\circ} - b) + (sin (b))^{2}

Rewrite using trig identities

= (cos (b))^{2} + (sin (b))^{2}

Remove parentheses:

(a) = a

= cos^{2} (b) + sin^{2} (b)

Rewrite using trig identities

= 1

We showed that the two sides could take the same form

\Rightarrow True

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

cos^{2} (x) = \frac{1 - tan ^{2} ( x )}{sec ^{2} ( x )}

sin^{2} (x) - 3 cos (x) - 4 = 0

tan^{2} (x) + sec (x) = 5

4 sin^{5} (x) = 3

Is sin^2(90-b)+cos^2(b-450)=1 ?
The answer to whether sin^2(90-b)+cos^2(b-450)=1 is True