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प्रश्न
Simplify : 2 sin30 + 3 tan45.
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उत्तर
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संबंधित प्रश्न
If `x/a=y/b = z/c` show that `x^3/a^3 + y^3/b^3 + z^3/c^3 = (3xyz)/(abc)`.
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`(1 - cos^2 A) cosec^2 A = 1`
Prove the following trigonometric identities.
`(1 + cos theta + sin theta)/(1 + cos theta - sin theta) = (1 + sin theta)/cos theta`
Prove the following trigonometric identities.
(sec A − cosec A) (1 + tan A + cot A) = tan A sec A − cot A cosec A
Prove the following identities:
(1 + cot A – cosec A)(1 + tan A + sec A) = 2
Prove that:
`(sinA - cosA)(1 + tanA + cotA) = secA/(cosec^2A) - (cosecA)/(sec^2A)`
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Write the value of `( 1- sin ^2 theta ) sec^2 theta.`
Write the value of `(1 - cos^2 theta ) cosec^2 theta`.
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What is the value of 9cot2 θ − 9cosec2 θ?
sec4 A − sec2 A is equal to
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
Simplify
sin A `[[sinA -cosA],["cos A" " sinA"]] + cos A[[ cos A" sin A " ],[-sin A" cos A"]]`
Prove the following identity :
`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`
Prove the following identity :
`cosA/(1 - tanA) + sin^2A/(sinA - cosA) = cosA + sinA`
Prove that `(cos θ)/(1 - sin θ) = (1 + sin θ)/(cos θ)`.
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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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If tan θ + sec θ = l, then prove that sec θ = `(l^2 + 1)/(2l)`.
