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प्रश्न
Prove that cos θ sin (90° - θ) + sin θ cos (90° - θ) = 1.
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उत्तर
LHS = cos θ sin (90° - θ) + sin θ cos (90° - θ)
= cos θ. cos θ + sin θ. sin θ
= cos2θ + sin 2θ
= 1
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities
`((1 + sin theta)^2 + (1 + sin theta)^2)/(2cos^2 theta) = (1 + sin^2 theta)/(1 - sin^2 theta)`
Prove the following trigonometric identities.
`(cos theta)/(cosec theta + 1) + (cos theta)/(cosec theta - 1) = 2 tan theta`
Prove the following identities:
`cotA/(1 - tanA) + tanA/(1 - cotA) = 1 + tanA + cotA`
Simplify : 2 sin30 + 3 tan45.
Prove the following identity :
`sec^2A + cosec^2A = sec^2Acosec^2A`
Prove the following identity :
`[1/((sec^2θ - cos^2θ)) + 1/((cosec^2θ - sin^2θ))](sin^2θcos^2θ) = (1 - sin^2θcos^2θ)/(2 + sin^2θcos^2θ)`
If sin θ + cos θ = `sqrt(3)`, then prove that tan θ + cot θ = 1.
Prove that cos2θ . (1 + tan2θ) = 1. Complete the activity given below.
Activity:
L.H.S. = `square`
= `cos^2θ xx square` ...`[1 + tan^2θ = square]`
= `(cos θ xx square)^2`
= 12
= 1
= R.H.S.
Which of the following is true for all values of θ (0° ≤ θ ≤ 90°)?
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`
