Question

If x lies in second quadrant and 3tanx+4=0, then find the value of

2cotx- 5cosx+sinx.

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Answer 1

Step-by-step explanation:

Since, angle A lies in the second quadrant hence 90∘<A<180∘⟹cosA<0,tanA<0,sinA>0

Given 3tanA+4=0⟹tanA=−43 Now, we know

cotA=1tanA=1−4/3=−34

cosA=1secA=−11+tan2A√=11+(−4/3)2√=−35

sinA=|tanA|1+tan2A√=|−4/3|1+(−4/3)2√=45

Now, substituting all the corresponding values in the given expression, one should get

2cotA−5cosA+sinA=2(−34)−5(−35)+45

=−32+3+45

=2310