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| Some Trigonometrical Identities |
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| 1. sin A = cos (90o - A) |
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2.  |
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| 3. tan A x tan (90o - A) = 1 |
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| 4. sin2 A + cos2 A = 1 |
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| 5. 1 + tan2 A = sec2 A |
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| 6. 1 + cot2 A = cosec2A |
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| Let us prove the above identities. |
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| Let DABC be a right-angled triangle with
ÐB = 90o. |
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| Let BC = a, AC = b and AB = c. |
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…(1) |
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| A + C = 90o |
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C = 90o - A |
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cos (90o - A) …(2) |
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From (1) and (2), |
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| sin A = cos (90o - A) |
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| sine of an angle = cosine of its complement. |
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| Similarly we can prove that cos A = sin (90o - A) |
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Sin A and cos A  |
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…(1) |
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also …(2) |
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From (1) and (2), |
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…(1) |
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| But A + C = 90o |
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C = 90o - A |
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tan (90o - A) …(2) |
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from (1) and (2), |
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tan A x tan (90o - A)  |
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tan A x tan (90o - A) = 1 |
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| We can use the above proof to show that tan (90o - A) = cot A |
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| a2 + c2 = b2 (by Pythagoras Theorem) |
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by b2  |
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But sin
and
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(sin A)2 + (cos A)2 = 1 |
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sin2A + cos2A = 1 |
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| a2 + c2 = b2 (by Pythagoras Theorem) |
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by c2  |
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But
and
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(tan A)2 + 1 = (sec A)2 |
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1 + tan2 A = sec2 A |
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| Again a2 + c2 = b2 (by Pythagoras Theorem) |
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by a2  |
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But cot A and cosec A  |
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1 + (cot A)2 = (cosec A)2 |
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1 + cot2 A = cosec2 A |
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Prove that  |
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L.H.S  |
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( sin2 A + cos2 A = 1) |
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| = 2 cosec2A = R.H.S. |
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Prove that  |
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L.H.S. |
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= R.H.S. |
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L.H.S. |
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| = cosec q - cot q = R.H.S. |
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Prove that  |
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L.H.S. |
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| = 2 sec q = R.H.S. |
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