Advertisements
Advertisements
प्रश्न
Evaluate the following: sin(35° + θ) - cos(55° - θ) - tan(42° + θ) + cot(48° - θ)
Advertisements
उत्तर
sin(35° + θ) - cos(55° - θ) - tan(42° + θ) + cot(48° - θ)
= sin[90° - (55° - θ)] - cos(55° - θ) - tan[90° - (48° - θ)] + cot(48° - θ)
= cos(55° - θ) - cos(55° - θ) - cot(48° - θ) + cot(48° - θ)
= 0.
APPEARS IN
संबंधित प्रश्न
Solve the following equation for A, if tan 3 A = 1
Calculate the value of A, if (tan A - 1) (cosec 3A - 1) = 0
Solve for x : 3 tan2 (2x - 20°) = 1
If θ = 15°, find the value of: cos3θ - sin6θ + 3sin(5θ + 15°) - 2 tan23θ
If `sqrt(3)` sec 2θ = 2 and θ< 90°, find the value of
cos 3θ
In a trapezium ABCD, as shown, AB ‖ DC, AD = DC = BC = 24 cm and ∠A = 30°. Find: length of AB
Express each of the following in terms of trigonometric ratios of angles between 0° and 45°: sin65° + cot59°
Evaluate the following: cot20° cot40° cot45° cot50° cot70°
If A, B and C are interior angles of ΔABC, prove that sin`(("A" + "B")/2) = cos "C"/(2)`
Prove the following: `(tan(90° - θ)cotθ)/("cosec"^2 θ)` = cos2θ
