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