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प्रश्न
Prove that:
`(sin(180^circ - theta)cos(90^circ + theta)tan(270^circ - theta)cot(360^circ - theta))/(sin(360^circ - theta)cos(360^circ + theta)sin(270^circ - theta)cosec (-theta))` = -1
बेरीज
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उत्तर
LHS = `(sin(180^circ - theta)cos(90^circ + theta)tan(270^circ - theta)cot(360^circ - theta))/(sin(360^circ - theta)cos(360^circ + theta)sin(270^circ - theta)cosec (-theta))`
`= ((sin theta)(- sin theta)(cot theta)(- cot theta))/((- sin theta)(cos theta)(- cos theta)(- cosec theta))`
`= (- sin theta xx cot theta cot theta)/(cos theta xx cos theta "cosec" theta)`
`= (- sin theta xx (cos theta)/(sin theta) xx (cos theta)/(sin theta))/(cos theta xx cos theta xx 1/(sin theta))`
`= - 1 xx (sin theta)/(sin theta)` = - 1 = RHS
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