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प्रश्न
If `sin theta = x , " write the value of cot "theta .`
उत्तर
`cot theta = cos theta / sin theta `
=` sqrt(1-sin^2 theta)/sin theta`
=`sqrt(1-x^2)/2`
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संबंधित प्रश्न
Prove the following identities:
`(i) (sinθ + cosecθ)^2 + (cosθ + secθ)^2 = 7 + tan^2 θ + cot^2 θ`
`(ii) (sinθ + secθ)^2 + (cosθ + cosecθ)^2 = (1 + secθ cosecθ)^2`
`(iii) sec^4 θ– sec^2 θ = tan^4 θ + tan^2 θ`
Prove the following trigonometric identities.
`1/(sec A - 1) + 1/(sec A + 1) = 2 cosec A cot A`
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(cosec A – sin A) (sec A – cos A) (tan A + cot A) = 1
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`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
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`(1-tan^2 theta)/(cot^2-1) = tan^2 theta`
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Define an identity.
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If \[sec\theta + tan\theta = x\] then \[tan\theta =\]
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Prove the following identity :
`sec^2A.cosec^2A = tan^2A + cot^2A + 2`
Prove the following identity :
`(1 + tan^2θ)sinθcosθ = tanθ`
If sec θ = `25/7`, then find the value of tan θ.
Prove that sin2 θ + cos4 θ = cos2 θ + sin4 θ.
Prove that sec2 (90° - θ) + tan2 (90° - θ) = 1 + 2 cot2 θ.
Prove that (cosec A - sin A)( sec A - cos A) sec2 A = tan A.
If A + B = 90°, show that sec2 A + sec2 B = sec2 A. sec2 B.
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S = `square`
= `square/sintheta + sintheta/costheta`
= `(cos^2theta + sin^2theta)/square`
= `1/(sintheta*costheta)` ......`[cos^2theta + sin^2theta = square]`
= `1/sintheta xx 1/square`
= `square`
= R.H.S
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`
