Advertisements
Advertisements
प्रश्न
Prove that:
`(sinthetasin(90^circ - theta))/cot(90^circ - theta) = 1 - sin^2theta`
Advertisements
उत्तर
L.H.S. = `(sinthetasin(90^@-theta))/cot(90^@-theta)`
= `(sinthetacostheta)/tantheta`
= `(sinthetacostheta)/(sintheta/costheta)`
= cos2θ
= 1 – sin2θ = R.H.S.
संबंधित प्रश्न
Without using trigonometric tables, evaluate the following:
`( i)\frac{\cos37^\text{o}}{\sin53^\text{o}}\text{ }(ii)\frac{\sin41^\text{o}}{\cos 49^\text{o}}(iii)\frac{\sin30^\text{o}17'}{\cos59^\text{o}\43'}`
A triangle ABC is right angles at B; find the value of`(secA.cosecC - tanA.cotC)/sinB`
Find the value of x, if tan x = `(tan60^circ - tan30^circ)/(1 + tan60^circ tan30^circ)`
Use tables to find cosine of 26° 32’
If 5 tan θ − 4 = 0, then the value of \[\frac{5 \sin \theta - 4 \cos \theta}{5 \sin \theta + 4 \cos \theta}\] is:
If 3 cos θ = 5 sin θ, then the value of
If A, B and C are interior angles of a triangle ABC, then \[\sin \left( \frac{B + C}{2} \right) =\]
tan 5° ✕ tan 30° ✕ 4 tan 85° is equal to
If tan θ = cot 37°, then the value of θ is
Prove the following:
tan θ + tan (90° – θ) = sec θ sec (90° – θ)
