Advertisements
Advertisements
प्रश्न
The value of sin ( \[{45}^° + \theta) - \cos ( {45}^°- \theta)\] is equal to
पर्याय
2 cos \[\theta\]
0
2 sin \[\theta\]
1
Advertisements
उत्तर
We know that,
\[\sin\left( 90 - \theta \right) = \cos\theta\]
So,
\[\sin\left( 45°+ \theta \right) = \cos\left[ 90 - \left( 45° + \theta \right) \right] = \cos\left( 45° - \theta \right)\]
\[\therefore \sin\left( 45°+ \theta \right) - \cos\left( 45°- \theta \right)\]
\[ = \cos\left( 45° - \theta \right) - \cos\left( 45° - \theta \right)\]
\[ = 0\]
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities
`cos theta/(1 - sin theta) = (1 + sin theta)/cos theta`
Prove the following trigonometric identities
`(1 + tan^2 theta)/(1 + cot^2 theta) = ((1 - tan theta)/(1 - cot theta))^2 = tan^2 theta`
Prove the following trigonometric identities.
`(sec A - tan A)/(sec A + tan A) = (cos^2 A)/(1 + sin A)^2`
Prove that `sqrt((1 + cos theta)/(1 - cos theta)) + sqrt((1 - cos theta)/(1 + cos theta)) = 2 cosec theta`
Prove the following identities:
(1 + cot A – cosec A)(1 + tan A + sec A) = 2
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
If tan A = n tan B and sin A = m sin B, prove that:
`cos^2A = (m^2 - 1)/(n^2 - 1)`
(i)` (1-cos^2 theta )cosec^2theta = 1`
`(sin theta+1-cos theta)/(cos theta-1+sin theta) = (1+ sin theta)/(cos theta)`
If `sec theta + tan theta = p,` prove that
(i)`sec theta = 1/2 ( p+1/p) (ii) tan theta = 1/2 ( p- 1/p) (iii) sin theta = (p^2 -1)/(p^2+1)`
Prove the following identity :
`sqrt((1 + cosA)/(1 - cosA)) = cosecA + cotA`
Prove the following identity :
`(1 + tan^2θ)sinθcosθ = tanθ`
If `asin^2θ + bcos^2θ = c and p sin^2θ + qcos^2θ = r` , prove that (b - c)(r - p) = (c - a)(q - r)
If sin θ = `1/2`, then find the value of θ.
Prove that (cosec A - sin A)( sec A - cos A) sec2 A = tan A.
Prove the following identities.
`(1 - tan^2theta)/(cot^2 theta - 1)` = tan2 θ
If sin θ + cos θ = a and sec θ + cosec θ = b , then the value of b(a2 – 1) is equal to
Prove that `"cot A"/(1 - cot"A") + "tan A"/(1 - tan "A")` = – 1
If sin A = `1/2`, then the value of sec A is ______.
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`
