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प्रश्न
(1 + sin A)(1 – sin A) is equal to ______.
पर्याय
cosec2A
sin2A
sec2A
cos2A
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उत्तर
(1 + sin A)(1 – sin A) is equal to cos2A.
Explanation:
(1 + sin A)(1 – sin A) = (1)2 – (sin A)2 ......{(a + b)(a – b) = (a2 – b2)}
= 1 – sin2A
= cos2A
संबंधित प्रश्न
Prove the following trigonometric identities.
`tan theta - cot theta = (2 sin^2 theta - 1)/(sin theta cos theta)`
Prove that: `sqrt((sec theta - 1)/(sec theta + 1)) + sqrt((sec theta + 1)/(sec theta - 1)) = 2 cosec theta`
Prove that:
`tanA/(1 - cotA) + cotA/(1 - tanA) = secA "cosec" A + 1`
Prove the following identities:
`cotA/(1 - tanA) + tanA/(1 - cotA) = 1 + tanA + cotA`
If 4 cos2 A – 3 = 0, show that: cos 3 A = 4 cos3 A – 3 cos A
If sinA + cosA = m and secA + cosecA = n , prove that n(m2 - 1) = 2m
Prove that : `1 - (cos^2 θ)/(1 + sin θ) = sin θ`.
Prove that sin2 5° + sin2 10° .......... + sin2 85° + sin2 90° = `9 1/2`.
If sin θ (1 + sin2 θ) = cos2 θ, then prove that cos6 θ – 4 cos4 θ + 8 cos2 θ = 4
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`
