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प्रश्न
(sec A + tan A) (1 − sin A) = ______.
विकल्प
sec A
sin A
cosec A
cos A
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उत्तर
(sec A + tan A) (1 − sin A) = cos A.
Explanation:
The given expression is `(sec "A"+tan "A") (1-sin "A")`.
Simplifying the given expression, we have
`(sec "A"+tan "A")(1-sin "A")`
= `(1/cos "A"+sin "A"/cos "A")(1-sin "A")`
= `(1+sin "A")/(cos"A")xx(1-sin "A")`
= `((1+sin "A")(1-sin "A"))/(cos "A")`
= `(1-sin^2 "A")/cos "A"`
= `cos^2 "A"/cos "A"`
= `cos "A"`
संबंधित प्रश्न
If sinθ + cosθ = p and secθ + cosecθ = q, show that q(p2 – 1) = 2p
if `cos theta = 5/13` where `theta` is an acute angle. Find the value of `sin theta`
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tan2θ cos2θ = 1 − cos2θ
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Prove that `(sec theta - 1)/(sec theta + 1) = ((sin theta)/(1 + cos theta))^2`
Prove the following identities:
`(1+ sin A)/(cosec A - cot A) - (1 - sin A)/(cosec A + cot A) = 2(1 + cot A)`
Prove that:
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If tan A = n tan B and sin A = m sin B, prove that `cos^2A = (m^2 - 1)/(n^2 - 1)`
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Prove the following identities:
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If cos θ = `24/25`, then sin θ = ?
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`
