Advertisements
Advertisements
प्रश्न
What is the value of 9cot2 θ − 9cosec2 θ?
उत्तर
We have,
`9 cot^2 θ-9 cosec^2θ= 9(cot ^2θ-cosec^2 θ) `
=` -9(cosec ^2θ-cot θ)`
We know that, `cosec ^2 θ-cot ^2 θ-1`
Therefore,
\[9 \cot^2 \theta - 9 {cosec}^2 \theta = - 9\]
APPEARS IN
संबंधित प्रश्न
Prove that `\frac{\sin \theta -\cos \theta }{\sin \theta +\cos \theta }+\frac{\sin\theta +\cos \theta }{\sin \theta -\cos \theta }=\frac{2}{2\sin^{2}\theta -1}`
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 identity.
`cos^2 A + 1/(1 + cot^2 A) = 1`
Prove the following trigonometric identities.
(1 + tan2θ) (1 − sinθ) (1 + sinθ) = 1
Prove the following identities:
`(1 + sin A)/(1 - sin A) = (cosec A + 1)/(cosec A - 1)`
Prove the following identities:
`cot^2A/(cosecA + 1)^2 = (1 - sinA)/(1 + sinA)`
`1+((tan^2 theta) cot theta)/(cosec^2 theta) = tan theta`
`(tan^2theta)/((1+ tan^2 theta))+ cot^2 theta/((1+ cot^2 theta))=1`
`cot theta/((cosec theta + 1) )+ ((cosec theta +1 ))/ cot theta = 2 sec theta `
If `cot theta = 1/ sqrt(3) , "write the value of" ((1- cos^2 theta))/((2 -sin^2 theta))`
What is the value of \[6 \tan^2 \theta - \frac{6}{\cos^2 \theta}\]
(cosec θ − sin θ) (sec θ − cos θ) (tan θ + cot θ) is equal to
If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan θ, then\[\frac{x^2}{a^2} + \frac{y^2}{b^2}\]
Prove the following identity :
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
Prove the following identity :
`tan^2A - tan^2B = (sin^2A - sin^2B)/(cos^2Acos^2B)`
Prove that `sin^2 θ/ cos^2 θ + cos^2 θ/sin^2 θ = 1/(sin^2 θ. cos^2 θ) - 2`.
Prove that `(sin θ. cos (90° - θ) cos θ)/sin( 90° - θ) + (cos θ sin (90° - θ) sin θ)/(cos(90° - θ)) = 1`.
Prove that `"cot A"/(1 - tan "A") + "tan A"/(1 - cot"A")` = 1 + tan A + cot A = sec A . cosec A + 1
Prove the following trigonometry identity:
(sinθ + cosθ)(cosecθ – secθ) = cosecθ.secθ – 2 tanθ
