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प्रश्न
Proved that cosec2(90° - θ) - tan2 θ = cos2(90° - θ) + cos2 θ.
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उत्तर
LHS = cosec2(90° - θ) - tan2 θ
LHS = sec2 θ - tan2 θ = 1
RHS = cos2(90° - θ) + cos2 θ
RHS = sin2θ + cos2 θ = 1
Hence, LHS = RHS
Hence proved.
संबंधित प्रश्न
(1 + tan θ + sec θ) (1 + cot θ − cosec θ) = ______.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(tan theta)/(1-cot theta) + (cot theta)/(1-tan theta) = 1+secthetacosectheta`
[Hint: Write the expression in terms of sinθ and cosθ]
Prove the following trigonometric identities.
`((1 + sin theta - cos theta)/(1 + sin theta + cos theta))^2 = (1 - cos theta)/(1 + cos theta)`
`(1 + cot^2 theta ) sin^2 theta =1`
`sin^6 theta + cos^6 theta =1 -3 sin^2 theta cos^2 theta`
Write the value of `sin theta cos ( 90° - theta )+ cos theta sin ( 90° - theta )`.
Prove that:
`sqrt(( secθ - 1)/(secθ + 1)) + sqrt((secθ + 1)/(secθ - 1)) = 2 "cosec"θ`
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
The value of 2sinθ can be `a + 1/a`, where a is a positive number, and a ≠ 1.
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`
