Advertisements
Advertisements
प्रश्न
Prove that:
sec2θ + cosec2θ = sec2θ x cosec2θ
Advertisements
उत्तर
L.H.S = sec2θ + cosec2θ
= 1 + tan2θ + 1 + cot2θ .....[∵ sec2θ = 1 + tan2θ and cosec2θ = 1 + cot2θ]
= 2 + tan2θ + cot2θ .....(i)
R.H.S = sec2θ x cosec2θ
= (1 + tan2θ) x (1 + cot2θ) .....[∵ sec2θ = 1 + tan2θ and cosec2θ = 1 + cot2θ]
= 1 + cot2θ + tan2θ + tan2θ x cot2θ
= 1 + cot2θ + tan2θ + tan2θ x (1/tan2θ) ...... [∵ cot2θ = 1/tan2θ]
= 2 + tan2θ + cot2θ .......(ii)
From (i) and (ii)
sec2θ + cosec2θ = sec2θ x cosec2θ
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities:
`(\text{i})\text{ }\frac{\sin \theta }{1-\cos \theta }=\text{cosec}\theta+\cot \theta `
Express the ratios cos A, tan A and sec A in terms of sin A.
Prove the following trigonometric identities.
`1/(sec A - 1) + 1/(sec A + 1) = 2 cosec A cot A`
Prove the following identities:
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
Prove the following identities:
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (1 + cosA)/sinA`
`1/((1+tan^2 theta)) + 1/((1+ tan^2 theta))`
`tan theta /((1 - cot theta )) + cot theta /((1 - tan theta)) = (1+ sec theta cosec theta)`
`(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos ^3 θ - sin^3 θ)/(cos θ - sin θ) = 2`
Prove the following identities:
`(sec theta + tan theta)/(sec theta - tan theta) = (sec theta + tan theta)^2 = 1 + 2 tan^2 theta + 2 sec theta tan theta`
Write True' or False' and justify your answer the following :
The value of the expression \[\sin {80}^° - \cos {80}^°\]
9 sec2 A − 9 tan2 A is equal to
Prove the following identity :
`sec^2A + cosec^2A = sec^2Acosec^2A`
Prove the following identity :
`cosecA + cotA = 1/(cosecA - cotA)`
Find the value of `θ(0^circ < θ < 90^circ)` if :
`tan35^circ cot(90^circ - θ) = 1`
Find the value of x , if `cosx = cos60^circ cos30^circ - sin60^circ sin30^circ`
Without using trigonometric identity , show that :
`sin(50^circ + θ) - cos(40^circ - θ) = 0`
If cosθ + sinθ = `sqrt2` cosθ, show that cosθ - sinθ = `sqrt2` sinθ.
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 cot2θ – tan2θ = cosec2θ – sec2θ.
Show that tan 7° × tan 23° × tan 60° × tan 67° × tan 83° = `sqrt(3)`.
