Advertisements
Advertisements
प्रश्न
Evaluate: Cosec (65 + θ) – sec (25 – θ) – tan (55 – θ) + cot (35 + θ)
उत्तर
cosec (65 + θ) = sec (90 – (65 + θ)) = sec (25 – θ)
tan (55 – θ) = cot (90 – (55 – θ) = cot (35 + θ)
⇒ sec (25 – θ) – sec (25 – θ) tan (55 – θ) + tan (55 – θ) = 0
APPEARS IN
संबंधित प्रश्न
If A, B and C are interior angles of a triangle ABC, then show that `\sin( \frac{B+C}{2} )=\cos \frac{A}{2}`
Evaluate the following :
`(sin 21^@)/(cos 69^@)`
Evaluate the following :
`(cot 40^@)/cos 35^@ - 1/2 [(cos 35^@)/(sin 55^@)]`
Without using trigonometric tables, prove that:
cos54° cos36° − sin54° sin36° = 0
Prove that
sin (70° + θ) − cos (20° − θ) = 0
Express each of the following in terms of trigonometric ratios of angles lying between 0° and 45°.
sec78° + cosec56°
If tan θ = cot θ and 0°∠θ ∠90°, state the value of θ
If A = 30°;
show that:
`(1 – cos 2"A")/(sin 2"A") = tan"A"`
Without using tables, evaluate the following: cosec245° sec230° - sin230° - 4cot245° + sec260°.
Prove the following:
`(sqrt(3) + 1) (3 - cot 30^circ)` = tan3 60° – 2 sin 60°