Evaluate the following: sec16° tan28° - cot62° cosec74°

बेरीज

sec16° tan28° - cot62° cosec74°
= sec(90° - 74°)tan(90° - 62°) - cot62° cosec74°
= cosec74° cot62° - cot62° cosec74°
= 0

shaalaa.com

Trigonometric Equation Problem and Solution

या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?

पाठ 27: Trigonometrical Ratios of Standard Angles - Exercise 27.3

पाठ 27 Trigonometrical Ratios of Standard Angles
Exercise 27.3 | Q 2.5

Solve the following equation for A, if sin 3 A = `sqrt3 /2`

If 4 sin² θ – 1 = 0 and angle θ is less than 90°, find the value of θ and hence the value of cos² θ + tan² θ.

Find the magnitude of angle A, if 2 sin A cos A - cos A - 2 sin A + 1 = 0

If 4 cos² x = 3 and x is an acute angle;
find the value of :
(i) x
(ii) cos² x + cot² x
(iii) cos 3x (iv) sin 2x

Calculate the value of A, if (cosec 2A - 2) (cot 3A - 1) = 0

Solve for x : sin (x + 10°) = `(1)/(2)`

Find the value of 'A', if cosec 3A = `(2)/sqrt(3)`

If A = B = 60°, verify that: sin(A - B) = sinA cosB - cosA sinB

Express each of the following in terms of trigonometric ratios of angles between 0° and 45°: cosec64° + sec70°

Evaluate the following: sin(35° + θ) - cos(55° - θ) - tan(42° + θ) + cot(48° - θ)