Some Trigonometrical Identities

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1. sin A = cos (90^o - A)

2.

3. tan A x tan (90^o - 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 = 90^o.

Let BC = a, AC = b and AB = c.

…(1)

A + C = 90^o

C = 90^o - A

cos (90^o - A) …(2)

From (1) and (2),

sin A = cos (90^o - A)

sine of an angle = cosine of its complement.

Similarly we can prove that cos A = sin (90^o - A)

Sin A and cos A

…(1)

also …(2)

From (1) and (2),

…(1)

But A + C = 90^o

C = 90^o - A

tan (90^o - A) …(2)

from (1) and (2),

tan A x tan (90^o - A)

tan A x tan (90^o - A) = 1

We can use the above proof to show that tan (90^o - 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.