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Complementary Trigonometric Ratios

tan 38∘- 22∘=...

Question

tan38°-cot22°=?

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Solution

LHS

=tan38°-cot22°

=(sin38°/cos38°)-(cos22°/sin22°)

=(sin38°sin22°-cos38°cos22°)/(cos38°)(sin22°)

(since sinAsinB-cosAcosB=-cos(A+B)

=-cos(38°+22°)(sec38°)(cosec22°)

=-cos60°cosec22°sec38°

=-½cosec22°sec38°.

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Similar questions

tan38° – cot22° = ?

(A)1/2 cosec38°sec22°

(B) 2sin22°cos38°

(C) –1/2 cosec22°sec38°

(D) none of these

Q. Without using trigonometric tables, evaluate that :

\frac{sec 39^{\circ}}{cos e c 51^{\circ}} + \frac{2}{\sqrt{3}} tan 17^{\circ} tan 38^{\circ} tan 60^{\circ} tan 52^{\circ} tan 73^{\circ} - 3 ({sin} 31^{\circ} + {sin}^{2} 59^{\circ})

Q. Without using trigonometric table evaluate the following:

\frac{sec 39^{\circ}}{cosec 51^{\circ}} + \frac{2}{\sqrt{3}} tan 17^{\circ} tan 38^{\circ} tan 60^{\circ} tan 52^{\circ} tan 73^{\circ} - 3 ({sin}^{2} 31^{\circ} + {sin}^{2} 59^{\circ})

Q. Without using trigonometric tables, evaluate:
(i)

\frac{\sin 50 °}{\cos 40 °} + \frac{cosec 40 °}{\sec 50 °} - 4 \cos 52 ° cosec 38 °

\frac{\cos 75 °}{\sin 15 °} + \frac{\sin 12 °}{\cos 78 °} - \cos 18 ° cosec 72 °

\frac{2 cosec 67 °}{\sec 23 °} - \frac{\tan 70 °}{\cot 20 °} - \cos 0 ° + \tan 38 ° \tan 52 °

\frac{\tan 76 °}{\cot 14 °} + \frac{\sec 58 °}{cosec 32 °} - \sin 35 ° \sec 55 ° - 8 \sin^{2} 30 °

Two persons are standing on the opposite sides of a tower. They observe the angles of elevation of the top of the tower to be $30^{\circ}$ and $38^{\circ}$ respectively. Find the distance between them, if the height of the tower is 50 m. [ $tan 38^{\circ} = 0.7813$ ] [2 MARKS]

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