Advertisements
Advertisements
प्रश्न
1 + cot2θ = ?
पर्याय
tan2θ
sec2θ
cosec2θ
cos2θ
Advertisements
उत्तर
1 + cot2θ = cosec2θ
Explanation:
`cot^2θ = (cos^2θ)/(sin^2θ)`
So, `1 + cot^2θ = (sin^2θ + cos^2θ)/(sin^2θ)`
= `1/(sin^2θ)`
= cosec2θ
APPEARS IN
संबंधित प्रश्न
Prove that `cosA/(1+sinA) + tan A = secA`
Prove the following identities:
`cosecA + cotA = 1/(cosecA - cotA)`
Prove the following identities:
`cot^2A/(cosecA + 1)^2 = (1 - sinA)/(1 + sinA)`
Prove the following identities:
`1 - cos^2A/(1 + sinA) = sinA`
If sin A + cos A = m and sec A + cosec A = n, show that : n (m2 – 1) = 2 m
`(1-cos^2theta) sec^2 theta = tan^2 theta`
` (sin theta + cos theta )/(sin theta - cos theta ) + ( sin theta - cos theta )/( sin theta + cos theta) = 2/ ((1- 2 cos^2 theta))`
`(sec theta + tan theta )/( sec theta - tan theta ) = ( sec theta + tan theta )^2 = 1+2 tan^2 theta + 25 sec theta tan theta `
`(cot^2 theta ( sec theta - 1))/((1+ sin theta))+ (sec^2 theta(sin theta-1))/((1+ sec theta))=0`
What is the value of \[6 \tan^2 \theta - \frac{6}{\cos^2 \theta}\]
If sinA + cosA = m and secA + cosecA = n , prove that n(m2 - 1) = 2m
Without using trigonometric table , evaluate :
`cos90^circ + sin30^circ tan45^circ cos^2 45^circ`
Find the value of ( sin2 33° + sin2 57°).
Prove that: 2(sin6θ + cos6θ) - 3 ( sin4θ + cos4θ) + 1 = 0.
Prove that `sin A/(sec A + tan A - 1) + cos A/(cosec A + cot A - 1) = 1`.
Prove that sin4A – cos4A = 1 – 2cos2A
If 2sin2θ – cos2θ = 2, then find the value of θ.
If a sinθ + b cosθ = c, then prove that a cosθ – b sinθ = `sqrt(a^2 + b^2 - c^2)`.
Prove that `(cot A - cos A)/(cot A + cos A) = (cos^2 A)/(1 + sin A)^2`
