Advertisements
Advertisements
प्रश्न
Prove the following trigonometric identities.
`((1 + sin theta - cos theta)/(1 + sin theta + cos theta))^2 = (1 - cos theta)/(1 + cos theta)`
Advertisements
उत्तर
In the given question, we need to prove `((1 + sin theta - cos theta)/(1 + sin theta + cos theta))^2 = (1 - cos theta)/(1 + cos theta)`
Taking `sin theta` common from the numerator and the denominator of the L.H.S, we get
`((1 + sin theta - cos theta)/(1 + sin theta + cos theta))^2 = (((sin theta)(cosec theta + 1 -cot theta))/((sin theta)(cosec theta + 1 + cot theta)))^2`
`= ((1 + cosec theta - cot theta)/(1 + cosec theta + cot theta))^2`
Now, using the property `1 + cot^2 theta = cosec^2 theta` we get
`((1 + cosec theta - cot theta)/(1 + cosec theta + cot theta))^2 = (((cosec^2 theta - cot^2 theta) +cosec theta - cot theta)/(1 + cosec theta + cot theta))^2`
Using `a^2 - b^2 = (a + b)(a - b) we get
`(((cosec^2 theta - cot^2 theta)(cosec theta - cot theta))/(1 + cosec theta + cot theta))^2 = (((cosec theta - cot theta)(cosec theta + cot theta + 1))/(1 + cosec theta + cot theta))^2`
`= (cosec theta - cot theta)^2`
Using `cot theta = cos theta/sin theta` and `cosec = 1/sin theta` we get
`(cosec theta - cot theta)^2 = (1/sin theta - cos theta/sin theta)^2`
`= ((1 - cos theta)/sin theta)^2`
Now, using the property `sin^2 theta + cos^2 theta = 1` we get
`(1 - cos theta)^2/sin^2 theta = (1 - cos theta)/(1 - cos^2 theta)`
`= (1 - cos theta)^2/((1 + cos theta)(1 - cos theta))`
`= (1 - cos theta)/(1 + cos theta)`
Hence proved.
APPEARS IN
संबंधित प्रश्न
Prove that sin6θ + cos6θ = 1 – 3 sin2θ. cos2θ.
Prove that:
sec2θ + cosec2θ = sec2θ x cosec2θ
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.
`cos A/(1 - tan A) + sin A/(1 - cot A) = sin A + cos A`
Prove the following trigonometric identities.
`1/(sec A + tan A) - 1/cos A = 1/cos A - 1/(sec A - tan A)`
Prove that
`sqrt((1 + sin θ)/(1 - sin θ)) + sqrt((1 - sin θ)/(1 + sin θ)) = 2 sec θ`
Prove the following identities:
`cot^2A((secA - 1)/(1 + sinA)) + sec^2A((sinA - 1)/(1 + secA)) = 0`
`((sin A- sin B ))/(( cos A + cos B ))+ (( cos A - cos B ))/(( sinA + sin B ))=0`
If tanθ `= 3/4` then find the value of secθ.
If x = r sin θ cos ϕ, y = r sin θ sin ϕ and z = r cos θ, then
If sin θ + sin2 θ = 1, then cos2 θ + cos4 θ =
Prove the following identity :
`(cosA + sinA)^2 + (cosA - sinA)^2 = 2`
Prove the following Identities :
`(cosecA)/(cotA+tanA)=cosA`
Prove that:
`sqrt(( secθ - 1)/(secθ + 1)) + sqrt((secθ + 1)/(secθ - 1)) = 2cosecθ`
Without using the trigonometric table, prove that
tan 10° tan 15° tan 75° tan 80° = 1
Prove that: sin4 θ + cos4θ = 1 - 2sin2θ cos2 θ.
If 1 – cos2θ = `1/4`, then θ = ?
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S = `square`
= `square/sintheta + sintheta/costheta`
= `(cos^2theta + sin^2theta)/square`
= `1/(sintheta*costheta)` ......`[cos^2theta + sin^2theta = square]`
= `1/sintheta xx 1/square`
= `square`
= R.H.S
Prove that `(1 + sec "A")/"sec A" = (sin^2"A")/(1 - cos"A")`
Show that tan4θ + tan2θ = sec4θ – sec2θ.
