How do you verify the identity $sin t csc (\frac{π}{2} - t) = tan t$ $sintcsc(pi/2-t)=tant$ ?

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Answer 1 · 2017-01-17T19:05:19

see below

Left Hand Side:

$sin t csc (\frac{π}{2} - t) = sin t \cdot \frac{1}{sin (\frac{π}{2} - t)}$ $sin t csc( pi/2 -t)=sint *1/(sin(pi/2 - t))$

Use the formula $sin (A - B) = sin A cos B - cos A sin B$ $sin(A-B)=sinAcosB-cosAsinB$

$= sin t \cdot \frac{1}{sin (\frac{π}{2}) cos t - cos (\frac{π}{2}) sin t}$ $=sin t * 1/(sin (pi/2) cos t-cos (pi/2) sin t)$

$= sin t \cdot \frac{1}{1 \cdot cos t - 0 \cdot sin t}$ $=sin t* 1/(1*cost-0*sint$

$= sin t \cdot \frac{1}{cos t}$ $=sint * 1/cos t$

$= \frac{sin t}{cos t}$ $= sint / cos t$

$= tan t$ $=tan t$

$:.=$ Right Hand Side

How do you verify the identity sintcsc(pi/2-t)=tantsintcsc(π2−t)=tantsintcsc(pi/2-t)=tant?