Advertisements
Advertisements
प्रश्न
Evaluate:
`(cos75^@)/(sin15^@) + (sin12^@)/(cos78^@) - (cos18^@)/(sin72^@)`
Advertisements
उत्तर
`(cos75^@)/(sin15^@) + (sin12^@)/(cos78^@) - (cos18^@)/(sin72^@)`
= `(cos(90^@ - 15^@))/(sin15^@) + (sin(90^@ - 78^@))/(cos78^@) - (cos(90^@ - 72^@))/(sin72^@)`
= `(sin15^@)/(sin15^@) + (cos78^@)/(cos78^@) - (sin72^@)/(sin72^@)`
= 1 + 1 – 1
= 1
संबंधित प्रश्न
Evaluate `(tan 26^@)/(cot 64^@)`
Express the following in terms of angles between 0° and 45°:
cos74° + sec67°
Find the value of x, if tan x = `(tan60^circ - tan30^circ)/(1 + tan60^circ tan30^circ)`
If A and B are complementary angles, prove that:
`(sinA + sinB)/(sinA - sinB) + (cosB - cosA)/(cosB + cosA) = 2/(2sin^2A - 1)`
If \[\frac{160}{3}\] \[\tan \theta = \frac{a}{b}, \text{ then } \frac{a \sin \theta + b \cos \theta}{a \sin \theta - b \cos \theta}\]
If 3 cos θ = 5 sin θ, then the value of
If A + B = 90°, then \[\frac{\tan A \tan B + \tan A \cot B}{\sin A \sec B} - \frac{\sin^2 B}{\cos^2 A}\]
Find the sine ratio of θ in standard position whose terminal arm passes through (4,3)
Prove that `(tan A)/(cot A) = (sec^2A)/("cosec"^2A)`.
Prove the following:
tan θ + tan (90° – θ) = sec θ sec (90° – θ)
