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`1/(c o s e ctheta-cottheta)-1/(sintheta)=1/(sintheta)-1/(c o s e ctheta+cottheta)` |
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Answer» We have `LHS=(1)/(("cosec" theta - cot theta))-(1)/(sin theta) ` `=(1)/(("cosec" theta - cot theta)) xx (("cosec" theta + cot theta))/(("cosec" theta + cot theta)) -(1)/(sin theta) ` ` =(("cosec" theta + cot theta))/(("cosec"^(2)theta - cot^(2)theta)) - "cosec" theta " "[because (1)/(sin theta)="cosec" theta] ` ` = ("cosec" theta + cot theta)- "cosec" theta " "[because "cosec"^(2)theta - cot^(2)theta =1 ] ` ` = cot theta. ` `RHS=(1)/(sin theta)-(1)/(("cosec" theta + cot theta))` `="cosec" theta - (1)/(("cosec" theta + cot theta))xx(("cosec" theta - cot theta))/(("cosec" theta - cot theta))` `= "cosec" theta - (("cosec" theta - cot theta))/(("cosec"^(2)theta - cot^(2)theta)) ` `="cosec" theta-("cosec" theta - cot theta) " " [ because "cosec"^(2)theta-cot^(2)theta=1] ` `= cot theta.` ` therefore LHS = RHS. ` |
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