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प्रश्न
Prove the following identities.
sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1
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उत्तर
sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1
L.H.S = sec6 θ
= (sec2 θ)3 = (1 + tan2 θ)3
= 1 + (tan2 θ)3 + 3 (1) (tan2 θ) (1 + tan2 θ) ......[(a + b)3 = a3 + b3 + 3 ab (a + b)]
= 1 + tan6 θ + 3 tan2 θ (1 + tan2 θ)
= 1 + tan6 θ + 3 tan2 θ (sec2 θ)
= 1 + tan6 θ + 3 tan2 θ sec2 θ
= tan6 θ + 3 tan2 θ sec2 θ + 1
L.H.S = R.H.S
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संबंधित प्रश्न
Prove the following trigonometric identities.
sin2 A cot2 A + cos2 A tan2 A = 1
Prove the following trigonometric identities.
`1/(sec A - 1) + 1/(sec A + 1) = 2 cosec A cot A`
Prove the following identities:
`((1 + tan^2A)cotA)/(cosec^2A) = tan A`
Prove that:
`cot^2A/(cosecA - 1) - 1 = cosecA`
`costheta/((1-tan theta))+sin^2theta/((cos theta-sintheta))=(cos theta+ sin theta)`
`cot theta/((cosec theta + 1) )+ ((cosec theta +1 ))/ cot theta = 2 sec theta `
If `sin theta = 1/2 , " write the value of" ( 3 cot^2 theta + 3).`
If x = r sinA cosB , y = r sinA sinB and z = r cosA , prove that `x^2 + y^2 + z^2 = r^2`
Prove that `(1 + sin θ)/(1 - sin θ) = (sec θ + tan θ)^2`.
If cot θ = `40/9`, find the values of cosec θ and sinθ,
We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
`(square + square)/square` = cosec2θ
`square/square` = cosec2θ ......[Taking root on the both side]
cosec θ = `41/9`
and sin θ = `1/("cosec" θ)`
sin θ = `1/square`
∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
