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Question
Prove that: `1/(cosec"A" - cot"A") - 1/sin"A" = 1/sin"A" - 1/(cosec"A" + cot"A")`
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Solution
`= 1/(cosectheta-cottheta)xx (cosectheta+cottheta)/(cosectheta+cottheta)-1/sintheta`
`= (cosectheta+cottheta)/(cosec^2theta-cot^2theta) - 1/sintheta`
`cosectheta+cottheta - 1/sintheta`
`1/sintheta+costheta/sintheta - 1/sintheta`
`1/sintheta+(costheta-1)/sinthetaxx(costheta+1)/(costheta+1)`
`1/sintheta+(cos^2theta-1)/((1+costheta)sintheta)`
`1/sintheta-(1-cos^2theta)/(sintheta(1+costheta))`
`1/sintheta - (sin2theta)/(sintheta(1+costheta))`
`1/sintheta-sintheta/(1+costheta)`
`1/sintheta - (sintheta/sintheta)/(1/sintheta+costheta/sintheta)`
`1/sintheta-1/(cosectheta+cottheta)`= RHS
RELATED QUESTIONS
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`(cot^2 A(sec A - 1))/(1 + sin A) = sec^2 A ((1 - sin A)/(1 + sec A))`
`sec theta (1- sin theta )( sec theta + tan theta )=1`
`sin theta (1+ tan theta) + cos theta (1+ cot theta) = ( sectheta+ cosec theta)`
`(tan A + tanB )/(cot A + cot B) = tan A tan B`
Show that none of the following is an identity:
(i) `cos^2theta + cos theta =1`
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`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (cosA + 1)/sinA`
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`(cot^2θ(secθ - 1))/((1 + sinθ)) = sec^2θ((1-sinθ)/(1 + secθ))`
Evaluate:
`(tan 65°)/(cot 25°)`
Prove the following identities.
`(sin "A" - sin "B")/(cos "A" + cos "B") + (cos "A" - cos "B")/(sin "A" + sin "B")`
If cosec θ + cot θ = p, then prove that cos θ = `(p^2 - 1)/(p^2 + 1)`
