Advertisements
Advertisements
प्रश्न
The value of the expression [cosec(75° + θ) – sec(15° – θ) – tan(55° + θ) + cot(35° – θ)] is ______.
पर्याय
– 1
0
1
`3/2`
उत्तर
The value of the expression [cosec(75° + θ) – sec(15° – θ) – tan(55° + θ) + cot(35° – θ)] is 0.
Explanation:
According to the question,
We have to find the value of the equation,
cosec(75° + θ) – sec(15° – θ) – tan(55° + θ) + cot(35° – θ)
= cosec[90° – (15° – θ)] – sec(15° – θ) – tan(55° + θ) + cot[90° – (55° + θ)]
Since, cosec(90° – θ) = sec θ
And cot(90° – θ) = tan θ
We get,
= sec(15° – θ) – sec(15° – θ) – tan(55° + θ) + tan(55° + θ)
= 0
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identity.
`(sin theta - cos theta + 1)/(sin theta + cos theta - 1) = 1/(sec theta - tan theta)`
Prove the following identities:
(cos A + sin A)2 + (cos A – sin A)2 = 2
`1/((1+tan^2 theta)) + 1/((1+ tan^2 theta))`
`(1+ cos theta + sin theta)/( 1+ cos theta - sin theta )= (1+ sin theta )/(cos theta)`
`(cot^2 theta ( sec theta - 1))/((1+ sin theta))+ (sec^2 theta(sin theta-1))/((1+ sec theta))=0`
Write the value of ` sec^2 theta ( 1+ sintheta )(1- sintheta).`
Write the value of tan10° tan 20° tan 70° tan 80° .
If \[\sin \theta = \frac{1}{3}\] then find the value of 9tan2 θ + 9.
If a cos θ + b sin θ = 4 and a sin θ − b sin θ = 3, then a2 + b2 =
Prove the following identity :
`(secA - 1)/(secA + 1) = (1 - cosA)/(1 + cosA)`
Prove the following identity :
`cosA/(1 - tanA) + sinA/(1 - cotA) = sinA + cosA`
Prove that tan2Φ + cot2Φ + 2 = sec2Φ.cosec2Φ.
Prove that (cosec A - sin A)( sec A - cos A) sec2 A = tan A.
Prove the following identities:
`1/(sin θ + cos θ) + 1/(sin θ - cos θ) = (2sin θ)/(1 - 2 cos^2 θ)`.
Prove that `(tan θ + sin θ)/(tan θ - sin θ) = (sec θ + 1)/(sec θ - 1)`
Prove that `"cosec" θ xx sqrt(1 - cos^2theta)` = 1
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
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
sin(45° + θ) – cos(45° – θ) is equal to ______.
Statement 1: sin2θ + cos2θ = 1
Statement 2: cosec2θ + cot2θ = 1
Which of the following is valid?
