Advertisements
Advertisements
प्रश्न
Prove the following identities.
cot θ + tan θ = sec θ cosec θ
Prove that: tan θ + cot θ = sec θ cosec θ
Advertisements
उत्तर
L.H.S. = cot θ + tan θ
L.H.S. = `costheta/sintheta + sintheta/costheta`
L.H.S. = `(cos^2theta + sin^2theta)/(sintheta costheta)`
[cos2 θ + sin2 θ = 1]
L.H.S. = `1/(sintheta costheta)`
Use Reciprocal Identities:
The expression can be split into `(1/sin θ) xx (1/cos θ)`.
`1/sin θ` = cosec θ
`1/cos θ` = sec θ
L.H.S. = cosec θ.sec θ
L.H.S. = sec θ.cosec θ
∴ L.H.S. = R.H.S.
APPEARS IN
संबंधित प्रश्न
If `sec alpha=2/sqrt3` , then find the value of `(1-cosecalpha)/(1+cosecalpha)` where α is in IV quadrant.
Evaluate sin25° cos65° + cos25° sin65°
Prove the following trigonometric identities.
`sin^2 A + 1/(1 + tan^2 A) = 1`
Prove the following trigonometric identities
cosec6θ = cot6θ + 3 cot2θ cosec2θ + 1
Prove the following identities:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
If tan A = n tan B and sin A = m sin B, prove that `cos^2A = (m^2 - 1)/(n^2 - 1)`
`1/((1+tan^2 theta)) + 1/((1+ tan^2 theta))`
`1/((1+ sin θ)) + 1/((1 - sin θ)) = 2 sec^2 θ`
`(sin theta +cos theta )/(sin theta - cos theta)+(sin theta- cos theta)/(sin theta + cos theta) = 2/((sin^2 theta - cos ^2 theta)) = 2/((2 sin^2 theta -1))`
Prove the following identities:
`(cot^2 theta (sec theta - 1))/((1 + sin theta)) + (sec^2 theta(sin theta - 1))/((1 + sec theta)) = 0`
If tanθ `= 3/4` then find the value of secθ.
Simplify : 2 sin30 + 3 tan45.
If \[\sin \theta = \frac{1}{3}\] then find the value of 2cot2 θ + 2.
If \[\sin \theta = \frac{1}{3}\] then find the value of 9tan2 θ + 9.
If `x/(a cosθ) = y/(b sinθ) "and" (ax)/cosθ - (by)/sinθ = a^2 - b^2 , "prove that" x^2/a^2 + y^2/b^2 = 1`
Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.
Prove that:
`(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos^3 θ - sin^3 θ)/(cos θ - sin θ) = 2`
If `tan θ = 7/24`, then to find value of cos θ complete the activity given below.
Activity:
sec2θ = 1 + `square` ...[Fundamental tri. identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square/576`
sec2θ = `square/576`
sec θ = `square`
cos θ = `square` ...`[cos theta = 1/sectheta]`
`sqrt((1 - cos^2theta) sec^2 theta) = tan theta`
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`
