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प्रश्न
`(1+tan^2theta)(1+cot^2 theta)=1/((sin^2 theta- sin^4theta))`
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उत्तर
LHS= `(1+tan^2theta)(1+cot^2 theta)`
=`sec^2 theta. cosec^2 theta (∵ sec^2 theta - tan^2 theta=1 and cosec^2 - cot^2 theta =1)`
=`1/(cos^2 theta. sin^theta)`
=` 1/((1-sin^2 theta ) sin^2 theta`
=`1/(sin^2theta-sin^4theta)`
==RHS
Hence, LHS = RHS
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संबंधित प्रश्न
Prove the following trigonometric identities
`((1 + sin theta)^2 + (1 + sin theta)^2)/(2cos^2 theta) = (1 + sin^2 theta)/(1 - sin^2 theta)`
Prove the following trigonometric identities.
`(cot A - cos A)/(cot A + cos A) = (cosec A - 1)/(cosec A + 1)`
Prove the following identities:
`(1 + sin A)/(1 - sin A) = (cosec A + 1)/(cosec A - 1)`
Prove the following identities:
(cos A + sin A)2 + (cos A – sin A)2 = 2
Prove that:
`tanA/(1 - cotA) + cotA/(1 - tanA) = secA "cosec" A + 1`
If sin A + cos A = m and sec A + cosec A = n, show that : n (m2 – 1) = 2 m
Prove the following identities:
sec4 A (1 – sin4 A) – 2 tan2 A = 1
Prove that:
`(sinA - cosA)(1 + tanA + cotA) = secA/(cosec^2A) - (cosecA)/(sec^2A)`
If x= a sec `theta + b tan theta and y = a tan theta + b sec theta ,"prove that" (x^2 - y^2 )=(a^2 -b^2)`
If tanθ `= 3/4` then find the value of secθ.
Prove the following identity :
`sec^2A.cosec^2A = tan^2A + cot^2A + 2`
Prove the following identity :
`sin^4A + cos^4A = 1 - 2sin^2Acos^2A`
Evaluate:
`(tan 65°)/(cot 25°)`
Prove that:
`sqrt((sectheta - 1)/(sec theta + 1)) + sqrt((sectheta + 1)/(sectheta - 1)) = 2cosectheta`
Prove that sin4θ - cos4θ = sin2θ - cos2θ
= 2sin2θ - 1
= 1 - 2 cos2θ
Prove the following identities.
sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1
Choose the correct alternative:
1 + cot2θ = ?
If tan θ = `9/40`, complete the activity to find the value of sec θ.
Activity:
sec2θ = 1 + `square` ......[Fundamental trigonometric identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square`
sec θ = `square`
Prove that
sin2A . tan A + cos2A . cot A + 2 sin A . cos A = tan A + cot A
(tan θ + 2)(2 tan θ + 1) = 5 tan θ + sec2θ.
