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рдкреНрд░рд╢реНрди
`(cot^2 theta ( sec theta - 1))/((1+ sin theta))+ (sec^2 theta(sin theta-1))/((1+ sec theta))=0`
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LHS= `(cot^2 theta ( sec theta - 1))/((1+ sin theta))+ (sec^2 theta(sin theta-1))/((1+ sec theta))`
=`(cos^2 theta/sin^2 theta(1/costheta-1))/((+ sin theta)) + (1/cos^2 theta(sin theta -1))/((1+ 1/cos theta))`
=`((cos^2 theta)/(sin^2 theta )((1- cos theta)/(cos theta)))/((1+sin theta))+ (((sin theta -1 ))/(cos ^2theta ))/(((cos theta + 1 )/(cos theta)))`
=`(cos^2 theta (1- cos theta))/(sin^2 theta cos theta (1+ sin theta))+ ((sin theta -1) cos theta)/((cos theta +1 ) cos^2 theta)`
=`(cos theta (1-cos theta))/((1- cos^2 theta)(1+ sin theta)) + ((sin theta -1)cos theta)/((costheta + 1 ) (1- sin^2 theta))`
=`(cos theta (1-cos theta))/((1- cos theta )( 1+ cos theta )(1+ sin theta)) + (-(1 sin theta ) cos theta)/((cos theta +1)(1-sin theta )(1+ sin theta))`
=`cos theta/((1+ cos theta )(1+ sin theta)) - cos theta/((cos theta +1)(1+ sin theta))`
= ЁЭЬГ
= RHS
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Prove the following trigonometric identities.
`tan theta + 1/tan theta` = sec θ.cosec θ
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`1/(tanA + cotA) = sinAcosA`
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tan (55° + x) = cot (35° – x)
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Prove that `(sintheta + "cosec" theta)/sin theta` = 2 + cot2θ
Prove that
`(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1) = (1 + cos "A")/"sin A"`
Simplify (1 + tan2θ)(1 – sinθ)(1 + sinθ)
`(cos^2 θ)/(sin^2 θ) - 1/(sin^2 θ)`, in simplified form, is ______.
Find the value of sin2θ + cos2θ
Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
