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рдкреНрд░рд╢реНрди
`(1+ cos theta - sin^2 theta )/(sin theta (1+ cos theta))= cot theta`
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рдЙрддреНрддрд░
LHS= `(1+ cos theta - sin^2 theta )/(sin theta (1+ cos theta)`
=` ((1+ cos theta )- (1-cos^2 theta))/(sin theta(1+ cos theta))`
=`(cos theta + cos^2 theta)/( sin theta ( 1+ cos theta))`
=`(cos theta ( 1+ cos theta ))/ ( sin theta ( 1+ cos theta))`
=`cos theta/ sin theta`
= cot ЁЭЬГ
= RHS
Hence, L.H.S. = R.H.S.
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рд╕рдВрдмрдВрдзрд┐рдд рдкреНрд░рд╢реНтАНрди
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Write the value of `3 cot^2 theta - 3 cosec^2 theta.`
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Sin4θ - cos4θ = 1 - 2cos2θ
Write the value of sin A cos (90° − A) + cos A sin (90° − A).
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`(tanθ + secθ - 1)/(tanθ - secθ + 1) = (1 + sinθ)/(cosθ)`
Prove the following identity :
`1/(tanA + cotA) = sinAcosA`
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Choose the correct alternative:
cos θ. sec θ = ?
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If tan θ – sin2θ = cos2θ, then show that sin2 θ = `1/2`.
`sqrt((1 - cos^2theta) sec^2 theta) = tan theta`
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`
