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प्रश्न
Simplify : 2 sin30 + 3 tan45.
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उत्तर
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संबंधित प्रश्न
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`(sin ^2 63^@ + sin^2 27^@)/(cos^2 17^@+cos^2 73^@)`
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(1 + cot2 A) sin2 A = 1
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`(tan^3 theta)/(1 + tan^2 theta) + (cot^3 theta)/(1 + cot^2 theta) = sec theta cosec theta - 2 sin theta cos theta`
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`sin A/(sec A + tan A - 1) + cos A/(cosec A + cot A + 1) = 1`
Prove the following trigonometric identities.
`(cot^2 A(sec A - 1))/(1 + sin A) = sec^2 A ((1 - sin A)/(1 + sec A))`
Prove the following identities:
`secA/(secA + 1) + secA/(secA - 1) = 2cosec^2A`
Prove the following identities:
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Prove that:
`(sinA - cosA)(1 + tanA + cotA) = secA/(cosec^2A) - (cosecA)/(sec^2A)`
`1/((1+tan^2 theta)) + 1/((1+ tan^2 theta))`
`(sec theta + tan theta )/( sec theta - tan theta ) = ( sec theta + tan theta )^2 = 1+2 tan^2 theta + 25 sec theta tan theta `
Show that none of the following is an identity:
`tan^2 theta + sin theta = cos^2 theta`
If `cos theta = 7/25 , "write the value of" ( tan theta + cot theta).`
9 sec2 A − 9 tan2 A is equal to
Prove the following identity:
`cosA/(1 + sinA) = secA - tanA`
Prove that ( 1 + tan A)2 + (1 - tan A)2 = 2 sec2A
Prove that `tan^3 θ/( 1 + tan^2 θ) + cot^3 θ/(1 + cot^2 θ) = sec θ. cosec θ - 2 sin θ cos θ.`
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
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