Is 1*sin(t)+(-1)*cos(t-pi/2)=0 ?

The answer to whether 1*sin(t)+(-1)*cos(t-pi/2)=0 is True

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prove 1sin(t)+(-1)cos(t-pi/2)=0

prove

1 \cdot sin (t) + (- 1) \cdot cos (t - \frac{π}{2}) = 0

True

Solution steps

1 \cdot sin (t) + (- 1) cos (t - \frac{π}{2}) = 0

Manipulating left side

1 sin (t) + (- 1) cos (t - \frac{π}{2})

Rewrite using trig identities

= 1 \cdot sin (t) + (- 1) sin (t)

1 \cdot sin (t) + (- 1) sin (t) = 0

= 0

We showed that the two sides could take the same form

\Rightarrow True

\frac{cot ( x )}{sec ( x ) - tan ( x )} = csc (x) + 1

csc (x) - cot (x) = \frac{1}{( csc ( x ) + cot ( x ) )}

\frac{csc ( x )}{cot ( x )} = \frac{1}{cos ( x )}

\frac{1 + tan ^{2} ( θ )}{tan ^{2} ( θ )} = csc^{2} (θ)

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

Is 1*sin(t)+(-1)*cos(t-pi/2)=0 ?
The answer to whether 1*sin(t)+(-1)*cos(t-pi/2)=0 is True