Prove the following: sin58° sec32° + cos58° cosec32° = 2

बेरीज

L.H.S.
= sin58° sec32° + cos58° cosec32°

= `sin(90° - 32°) xx (1)/(cos32°) + cos(90° - 32°) xx (1)/(sin32°)`

= `cos32° xx (1)/(cos32°) + sin32° xx (1)/(sin32°)`

= 1 + 1
= 2
= R.H.S.

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Trigonometric Equation Problem and Solution

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पाठ 22: Trigonometrical Ratios of Standard Angles - Exercise 27.3

पाठ 22 Trigonometrical Ratios of Standard Angles
Exercise 27.3 | Q 14.2

If 4 cos² x° - 1 = 0 and 0 ∠ x° ∠ 90°,
find:(i) x°
(ii) sin² x° + cos² x°
(iii) `(1)/(cos^2xx°) – (tan^2 xx°)`

Solve the following equation for A, if 2 sin A = 1

Find the magnitude of angle A, if 2 cos² A - 3 cos A + 1 = 0

Solve for x : tan² (x - 5°) = 3

Find the value of 'A', if 2 sin 2A = 1

If θ = 15°, find the value of: cos3θ - sin6θ + 3sin(5θ + 15°) - 2 tan²3θ

If A = B = 60°, verify that: tan(A - B) = `(tan"A" - tan"B")/(1 + tan"A" tan"B"")`

Find the value of: `sqrt((1 - sin^2 60°)/(1 + sin^2 60°)` If 3 tan²θ - 1 = 0, find the value
a. cosθ
b. sinθ

Evaluate the following: `(3sin37°)/(cos53°) - (5"cosec"39°)/(sec51°) + (4tan23° tan37° tan67° tan53°)/(cos17° cos67° "cosec"73° "cosec"23°)`

If P, Q and R are the interior angles of ΔPQR, prove that `cot(("Q" + "R")/2) = tan "P"/(2)`