Advertisements
Advertisements
प्रश्न
`5/(sin^2theta) - 5cot^2theta`, complete the activity given below.
Activity:
`5/(sin^2theta) - 5cot^2theta`
= `square (1/(sin^2theta) - cot^2theta)`
= `5(square - cot^2theta) ......[1/(sin^2theta) = square]`
= 5(1)
= `square`
Advertisements
उत्तर
`5/(sin^2theta) - 5cot^2theta`
= `5 (1/(sin^2theta) - cot^2theta)`
= `5("cosec"^2theta - cot^2theta) ......[1/(sin^2theta) = "cosec"^2theta]`
= 5(1)
= 5.
APPEARS IN
संबंधित प्रश्न
If m=(acosθ + bsinθ) and n=(asinθ – bcosθ) prove that m2+n2=a2+b2
Evaluate
`(sin ^2 63^@ + sin^2 27^@)/(cos^2 17^@+cos^2 73^@)`
If `x/a=y/b = z/c` show that `x^3/a^3 + y^3/b^3 + z^3/c^3 = (3xyz)/(abc)`.
Prove the following trigonometric identities.
tan2 A sec2 B − sec2 A tan2 B = tan2 A − tan2 B
Prove the following identities:
`sqrt((1 + sinA)/(1 - sinA)) = cosA/(1 - sinA)`
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
`(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos ^3 θ - sin^3 θ)/(cos θ - sin θ) = 2`
`(sin theta)/((sec theta + tan theta -1)) + cos theta/((cosec theta + cot theta -1))=1`
`(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))`
`(cos theta cosec theta - sin theta sec theta )/(costheta + sin theta) = cosec theta - sec theta`
Write the value of`(tan^2 theta - sec^2 theta)/(cot^2 theta - cosec^2 theta)`
Prove that:
`(sin^2θ)/(cosθ) + cosθ = secθ`
If \[\cos A = \frac{7}{25}\] find the value of tan A + cot A.
Prove the following identity :
`(sinA + cosA)/(sinA - cosA) + (sinA - cosA)/(sinA + cosA) = 2/(2sin^2A - 1)`
Prove the following identity :
`(tanθ + 1/cosθ)^2 + (tanθ - 1/cosθ)^2 = 2((1 + sin^2θ)/(1 - sin^2θ))`
Prove that:
tan (55° + x) = cot (35° – x)
Prove that cosec2 (90° - θ) + cot2 (90° - θ) = 1 + 2 tan2 θ.
Prove that `sqrt((1 + cos A)/(1 - cos A)) = (tan A + sin A)/(tan A. sin A)`
Prove that: `1/(sec θ - tan θ) = sec θ + tan θ`.
If 5 sec θ – 12 cosec θ = 0, then find values of sin θ, sec θ
