Advertisements
Advertisements
प्रश्न
Evaluate: `(cot^2 41°)/(tan^2 49°) - 2 (sin^2 75°)/(cos^2 15°)`
Advertisements
उत्तर
`(cot^2 41°)/(tan^2 49°) - 2 (sin^2 75°)/(cos^2 15°)`
tan(49°) = cot(41°)
(because 49° + 41° = 90°)
`(cot^2 41°)/(tan^2 49°) = (cot^2 41°)/(cot41°)^2 = 1`
`(sin^2 75°)/(cos^2 15°)`
sin75° = cos15°
`(sin^2 75°)/(cos^2 15°) = (cos15°)^2/(cos15°)^2 = 1`
1 − 2(1) = 1 − 2
= −1
संबंधित प्रश्न
if `3 cos theta = 1`, find the value of `(6 sin^2 theta + tan^2 theta)/(4 cos theta)`
Prove that:
`(sinthetasin(90^circ - theta))/cot(90^circ - theta) = 1 - sin^2theta`
Use tables to find cosine of 26° 32’
Use trigonometrical tables to find tangent of 17° 27'
Use tables to find the acute angle θ, if the value of cos θ is 0.9848
Evaluate:
`sec26^@ sin64^@ + (cosec33^@)/sec57^@`
Prove that:
`1/(1 + sin(90^@ - A)) + 1/(1 - sin(90^@ - A)) = 2sec^2(90^@ - A)`
If \[\tan \theta = \frac{4}{5}\] find the value of \[\frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta}\]
If \[\tan A = \frac{3}{4} \text{ and } A + B = 90°\] then what is the value of cot B?
If \[\frac{160}{3}\] \[\tan \theta = \frac{a}{b}, \text{ then } \frac{a \sin \theta + b \cos \theta}{a \sin \theta - b \cos \theta}\]
