Complete the identity: csc(t)(sin(t) + cos(t)) = ?

Keeping in that csc(t) = 1/sin(t), distribute that csc(t) and you get csc(t)(sin(t) + cos(t)) = 1 + cot(t)

Complete the identity: csc(t)(sin(t) + cos(t)) = ?

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Whenever you have a complicated problem like this, it is best to break it down into Sines and Cosines- so

*csc(t)(sin(t)+cos(t)) * since csc(t) = 1/sin(t), we have

*(sin(t)+cos(t))*1/(sin(t)* Distributing out, we get

*sin(t)/sin(t) + cos(t)/sin(t) *This simplifies to

*1 + cot(t)*

George C. | Humboldt State and Georgetown graduateHumboldt State and Georgetown graduate

You have a typo here.

Recall #1 cst t =1/sin t , cst t is the inverse of sin t

#2 cos t = cos(t/2 + t/2)=cos t/2 cos t/2 - sin t/2 sin t/2 using the formula for cos(a+b)

where a=b=t/2

#3 1 = cos t/2 cos t/2 + sin t/2 sin t/2 , The fundamental relationship in Trigonometry

rewritten with minor acceptable changes.

Therefore, csc t sin t + cos t = [1/(sin t/2)] sin t/2 + [cos t/2 cos t/2 - sin t/2 sin t/2]

= 1 + cos t/2 cos t/2 - sin t/2 sin t/2

= cos t/2 cos t/2 + sin t/2 sin t/2 + cos t/2 cos t/2 - sin t/2 sin t/2

= 2 [ cos t/2 cos t/2 ]

= 2 [ cos t/2 ]^2 , read *2 multiplied by (cos t/2) squarred!*

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