Trigonometric functions

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#1
Can someone please help me work out question 6 (I) I keep trying to simplify the left hand side yet I canβt reach the answer on the right.
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2 years ago
#2
(Original post by Hannahkisley)
Can someone please help me work out question 6 (I) I keep trying to simplify the left hand side yet I canβt reach the answer on the right.
1. Expand the LHS.
2. Since the RHS is entirely in terms of sec and cosec, it would be good to turn the terms in the LHS entirely in terms of these as well. So replace tan^2 = sec^2 - 1 and cot^2 = cosec^2 - 1
3. Simplify.
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#3
(Original post by RDKGames)
1. Expand the LHS.
2. Since the RHS is entirely in terms of sec and cosec, it would be good to turn the terms in the LHS entirely in terms of these as well. So replace tan^2 = sec^2 - 1 and cot^2 = cosec^2 - 1
3. Simplify.
This is what I have got so far and donβt see how it can simplify to the answer, could you tell me where Iβve gone wrong
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2 years ago
#4
(Original post by Hannahkisley)
This is what I have got so far and donβt see how it can simplify to the answer, could you tell me where Iβve gone wrong
What you've posted looks correct. Try rewriting -2sec^2(x) as -2/cos^2(x), and -2cosec^2(x) as -2/sin^2(x), then combine those two terms over their common denominator.
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2 years ago
#5
(Original post by Hannahkisley)
This is what I have got so far and donβt see how it can simplify to the answer, could you tell me where Iβve gone wrong
Looks good. And it's almost the answer.

In fact you can split it

and you would arrive at their answer if you can show that this last bracket term which can be begun by saying that the LHS is and one more step would be enough to show its =0.
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