Evaluate the following: cot27° - tan63°

योग

cot27° - tan63°
= cot(90° - 63°) - tan63°
= tan63° - tan63°
= 0.

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Trigonometric Equation Problem and Solution

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अध्याय 22: Trigonometrical Ratios of Standard Angles - Exercise 27.3

अध्याय 22 Trigonometrical Ratios of Standard Angles
Exercise 27.3 | Q 2.2

In ΔABC, ∠B = 90° , AB = y units, BC = `(sqrt3)` units, AC = 2 units and angle A = x°, find:

Solve for x : cos `(x)/(3) –1` = 0

Solve for x : cos (2x - 30°) = 0

If `sqrt(3)` sec 2θ = 2 and θ< 90°, find the value of

cos² (30° + θ) + sin² (45° - θ)

Find the value 'x', if:

In the given figure; ∠B = 90°, ∠ADB = 30°, ∠ACB = 45° and AB = 24 m. Find the length of CD.

Evaluate the following: cos39° cos48° cos60° cosec42° cosec51°

If cos3θ = sin(θ - 34°), find the value of θ if 3θ is an acute angle.

If A, B and C are interior angles of ΔABC, prove that sin`(("A" + "B")/2) = cos "C"/(2)`

If A + B = 90°, prove that `(tan"A" tan"B" + tan"A" cot"B")/(sin"A" sec"B") - (sin^2"B")/(cos^2"A")` = tan²A