Prove that:
(sin A /cot A+ cosec A) = 2 +(sin A /cotA-cosec A)
taking (sin A /cotA-cosec A) to the lhs
(sin A /cot A+ cosec A)-(sin A /cotA-cosec A) =2
take LCM,you get
-2 cosecAsinA/-1 =2 (since cot2A-cosec2A)
(sin A /cot A+ cosec A)-(sin A /cotA-cosec A) =2
take LCM,you get
-2 cosecAsinA/-1 =2 (since cot2A-cosec2A)
But, can we actually take it to the LHS?
Is there a way to solve this question without taking it to the LHS?
Is there a way to solve this question without taking it to the LHS?
shine ali, sorry i am too late.
i did not notice your question.
yet there is another possible method,
take LCM on both sides you will get LHS=RHS=1-costheta
i did not notice your question.
yet there is another possible method,
take LCM on both sides you will get LHS=RHS=1-costheta
Have a look at page 2 below or here .
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