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प्रश्न
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`
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उत्तर
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = AC2 .....(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" = bbunderline(sintheta) and "BC"/"AC" = bbunderline(costheta)`
∴ `sin^2 theta + cos^2 theta = bbunderline1`
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संबंधित प्रश्न
Prove that `cosA/(1+sinA) + tan A = secA`
Prove the following trigonometric identity.
`cos^2 A + 1/(1 + cot^2 A) = 1`
Prove the following trigonometric identities.
`tan theta/(1 - cot theta) + cot theta/(1 - tan theta) = 1 + tan theta + cot theta`
Prove the following identities:
`((1 + tan^2A)cotA)/(cosec^2A) = tan A`
Prove that:
`(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA) = 2`
If` (sec theta + tan theta)= m and ( sec theta - tan theta ) = n ,` show that mn =1
If `(cot theta ) = m and ( sec theta - cos theta) = n " prove that " (m^2 n)(2/3) - (mn^2)(2/3)=1`
Write the value of `(1 + cot^2 theta ) sin^2 theta`.
If tan A =` 5/12` , find the value of (sin A+ cos A) sec A.
Prove that `(sinθ - cosθ + 1)/(sinθ + cosθ - 1) = 1/(secθ - tanθ)`
Prove the following identity :
`(1 - cos^2θ)sec^2θ = tan^2θ`
For ΔABC , prove that :
`tan ((B + C)/2) = cot "A/2`
Evaluate:
sin2 34° + sin2 56° + 2 tan 18° tan 72° – cot2 30°
Prove that: (1+cot A - cosecA)(1 + tan A+ secA) =2.
If cosθ + sinθ = `sqrt2` cosθ, show that cosθ - sinθ = `sqrt2` sinθ.
Prove that sin2 5° + sin2 10° .......... + sin2 85° + sin2 90° = `9 1/2`.
Prove that : `tan"A"/(1 - cot"A") + cot"A"/(1 - tan"A") = sec"A".cosec"A" + 1`.
If x = a tan θ and y = b sec θ then
Prove that `costheta/(1 + sintheta) = (1 - sintheta)/(costheta)`
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
