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प्रश्न
If sec θ + tan θ = x, then sec θ =
विकल्प
\[\frac{x^2 + 1}{x}\]
\[\frac{x^2 + 1}{2x}\]
\[\frac{x^2 - 1}{2x}\]
\[\frac{x^2 - 1}{x}\]
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उत्तर
Given: `sec θ+tan θ=1`
We know that,
`sec^2θ-tan^2θ=1`
⇒ `(secθ+tan θ)(secθ-tan θ)=1`
⇒`x(sec θ-tan θ)=1`
⇒ `secθ-tan θ=1/x`
Now,
`sec θ+tan =x`
`sec θ-tan θ=1/x`
Adding the two equations, we get
`(sec θ+tan θ)+(sec θ-tan θ)=x+1/x`
⇒` sec θ+tan θ+sec θ-tan θ=(x^2+1)/x`
⇒ `2 sec θ=(x^2+1)/x`
⇒` sec θ=(x^2+1)/(2x)`
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संबंधित प्रश्न
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`(1 + sin θ)/cos θ+ cos θ/(1 + sin θ) = 2 sec θ`
Prove the following trigonometric identity.
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Prove the following trigonometric identities.
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Prove the following trigonometric identities.
if x = a cos^3 theta, y = b sin^3 theta` " prove that " `(x/a)^(2/3) + (y/b)^(2/3) = 1`
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(1 + cos α) (1 + cos β) (1 + cos γ) = (1 − cos α) (1 − cos α) (1 − cos β) (1 − cos γ)
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`(1-cos^2theta) sec^2 theta = tan^2 theta`
`cos^2 theta + 1/((1+ cot^2 theta )) =1`
`sqrt((1+cos theta)/(1-cos theta)) + sqrt((1-cos theta )/(1+ cos theta )) = 2 cosec theta`
`((sin A- sin B ))/(( cos A + cos B ))+ (( cos A - cos B ))/(( sinA + sin B ))=0`
Prove the following identity :
secA(1 - sinA)(secA + tanA) = 1
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Prove the following identities.
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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`
