Question

Prove that sin²Q + tan²Q + cos²Q = sec²Q

Answer 1

Solution:

To prove: sin²Q + tan²Q + cos²Q = sec²Q

We can prove it by both sides.

Considering L.H.S:

= sin²Q + tan²Q + cos²Q

= sin²Q + cos²Q + tan²Q

= 1 + tan²Q

= sec²Q

= R.H.S

Considering R.H.S:

= sec²Q

= 1 + tan²Q

= sin²Q + cos²Q + tan²Q

= sin²Q + tan²Q + cos²Q

= L.H.S

Trigonometric identities used:

• sin²Q + cos²Q = 1
• 1 + tan²Q = sec²Q

More Trigonometric identities:

• 1 - cos²Q = sin²Q
• 1 - sin²Q = cos²Q
• sec²Q - tan²Q = 1
• sec²Q - 1 = tan²Q
• cosec²Q - cot²Q = 1
• cosec²Q - 1 = cot²Q
• cot²Q + 1 = cosec²Q

Answer 2

Given

• sin²Q + tan²Q + cos²Q = sec²Q

To prove

• sin²Q + tan²Q + cos²Q = sec²Q

Solution

LHS :-

• sin²Q + tan²Q + cos²Q (sin²Q + cos²Q = 1)
• 1 +  tan²Q (1 + tan²Q = ²Q)
• sec²Q

RHS :-

• sec²Q
• 1 + tan²Q
• sin²Q + tan²Q + cos²Q

Therefore, L.H.S = R.H.S