Advertisements
Advertisements
प्रश्न
If tan θ – sin2θ = cos2θ, then show that `sin^2θ = 1/2`.
Advertisements
उत्तर
tan θ – sin2θ = cos2θ ...[Given]
∴ tan θ = sin2θ + cos2θ
∴ tan θ = 1 ...[∵ sin2θ + cos2θ = 1]
But, tan 45° = 1
∴ tan θ = tan 45°
∴ θ = 45°
sin2θ = sin245°
= `(1/sqrt(2))^2`
= `1/2`
APPEARS IN
संबंधित प्रश्न
Without using trigonometric tables evaluate
`(sin 35^@ cos 55^@ + cos 35^@ sin 55^@)/(cosec^2 10^@ - tan^2 80^@)`
If 3 sin θ + 5 cos θ = 5, prove that 5 sin θ – 3 cos θ = ± 3.
If sin θ + cos θ = x, prove that `sin^6 theta + cos^6 theta = (4- 3(x^2 - 1)^2)/4`
`1+((tan^2 theta) cot theta)/(cosec^2 theta) = tan theta`
`(tan theta)/((sec theta -1))+(tan theta)/((sec theta +1)) = 2 sec theta`
`cot theta/((cosec theta + 1) )+ ((cosec theta +1 ))/ cot theta = 2 sec theta `
` (sin theta + cos theta )/(sin theta - cos theta ) + ( sin theta - cos theta )/( sin theta + cos theta) = 2/ ((1- 2 cos^2 theta))`
Find the value of `(cos 38° cosec 52°)/(tan 18° tan 35° tan 60° tan 72° tan 55°)`
\[\frac{1 - \sin \theta}{\cos \theta}\] is equal to
The value of (1 + cot θ − cosec θ) (1 + tan θ + sec θ) is
Prove the following identity :
`sinA/(1 + cosA) + (1 + cosA)/sinA = 2cosecA`
Prove the following identity :
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
Prove that:
`(cot A - 1)/(2 - sec^2 A) = cot A/(1 + tan A)`
Prove that: (1+cot A - cosecA)(1 + tan A+ secA) =2.
Prove that `sin^2 θ/ cos^2 θ + cos^2 θ/sin^2 θ = 1/(sin^2 θ. cos^2 θ) - 2`.
Without using a trigonometric table, prove that
`(cos 70°)/(sin 20°) + (cos 59°)/(sin 31°) - 8sin^2 30° = 0`.
`(1 - tan^2 45^circ)/(1 + tan^2 45^circ)` = ?
If cos A = `(2sqrt(m))/(m + 1)`, then prove that cosec A = `(m + 1)/(m - 1)`.
(1 – cos2 A) is equal to ______.
(1 + sin A)(1 – sin A) is equal to ______.
