Advertisements
Advertisements
प्रश्न
sin(45° + θ) – cos(45° – θ) is equal to ______.
पर्याय
2cosθ
0
2sinθ
1
Advertisements
उत्तर
sin(45° + θ) – cos(45° – θ) is equal to 0.
Explanation:
sin(45° + θ) – cos(45° – θ)
= cos[90° – (45° + θ)] – cos(45° – θ) ...[∵ cos(90° – θ) = sinθ]
= cos(45° – θ) – cos(45° – θ)
= 0
APPEARS IN
संबंधित प्रश्न
If acosθ – bsinθ = c, prove that asinθ + bcosθ = `\pm \sqrt{a^{2}+b^{2}-c^{2}`
Prove the following trigonometric identities.
sin2 A cot2 A + cos2 A tan2 A = 1
Prove the following identities:
(cosec A – sin A) (sec A – cos A) (tan A + cot A) = 1
`sqrt((1-cos theta)/(1+cos theta)) = (cosec theta - cot theta)`
Write the value of `(1 + cot^2 theta ) sin^2 theta`.
Write True' or False' and justify your answer the following :
The value of sin θ+cos θ is always greater than 1 .
Prove the following identity :
`(tanθ + secθ - 1)/(tanθ - secθ + 1) = (1 + sinθ)/(cosθ)`
Prove the following identity :
`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`
Prove the following identity :
`(cotA - cosecA)^2 = (1 - cosA)/(1 + cosA)`
Prove the following identity :
`cosA/(1 - tanA) + sin^2A/(sinA - cosA) = cosA + sinA`
Prove the following identity :
`1/(cosA + sinA - 1) + 2/(cosA + sinA + 1) = cosecA + secA`
Prove the following identity :
`2(sin^6θ + cos^6θ) - 3(sin^4θ + cos^4θ) + 1 = 0`
If tanA + sinA = m and tanA - sinA = n , prove that (`m^2 - n^2)^2` = 16mn
Find the value of `θ(0^circ < θ < 90^circ)` if :
`tan35^circ cot(90^circ - θ) = 1`
If sec θ = `25/7`, then find the value of tan θ.
Prove that `sqrt((1 - sin θ)/(1 + sin θ)) = sec θ - tan θ`.
Prove that `cot^2 "A" [(sec "A" - 1)/(1 + sin "A")] + sec^2 "A" [(sin"A" - 1)/(1 + sec"A")]` = 0
Prove that cos2θ . (1 + tan2θ) = 1. Complete the activity given below.
Activity:
L.H.S = `square`
= `cos^2theta xx square .....[1 + tan^2theta = square]`
= `(cos theta xx square)^2`
= 12
= 1
= R.H.S
Show that tan4θ + tan2θ = sec4θ – sec2θ.
If tan θ + sec θ = l, then prove that sec θ = `(l^2 + 1)/(2l)`.
