Question

Prove that = (1-cos²A).sec²B + tan²B (1-sin²A)=sin²A + tan²B​

Answer 1

Required Answer:-

Correct Question:

• Prove that (1 - cos²x).sec²x + tan²x.(1 - sin²x) = sin²x + tan²x

Proof:

To prove this, we need to prove LHS = RHS

We know that,

→ sin²x + cos²x = 1

→ sin²x = 1 - cos²x and cos²x = 1 - sin²x

→ tan x = sin x/cos x

→ sec x = 1/cos x

• Taking LHS:

= (1 - cos²x).sec²x + tan²x.(1 - sin²x)

= sin²x.sec²x + tan²x.cos²x [sin²x = 1 - cos²x and cos²x = 1 - sin²x]

= sin²x/cos²x + tan²x.cos²x [sec x = 1/cos x]

= tan²x + sin²x/cos²x × cos²x [sin x/cos x = tan x]

= tan²x + sin²x

= RHS •

Therefore, LHS = RHS (Proved)

Basic Trigonometry Formula:

1. Relationship between sides.

• sin x = Height/Hypotenuse.
• cos x = Base/Hypotenuse.
• tan x = Height/Base.
• cot x = Base/Height.
• sec x = Hypotenuse/Base.
• cosec x = Hypotenuse/Height.

2. Reciprocal Relation.

• sin x = 1/cosec x and cosec x = 1/sin x.
• cos x = 1/sec x and sec x = 1/cos x.
• tan x = 1/cot x and cot x = 1/tan x.

3. Cofunction Identities.

• sin(90° - x) = cos x and cos(90° - x) = sin(x)
• cosec(90° - x) = sec x and sec(90° - x) = cosec(x)
• tan(90° - x) = cot x and cot(90° - x) = tan x.

4. Pythagorean Identities.

• sin²x + cos²x = 1
• cosec²x - cot²x = 1
• sec²x - tan²x = 1

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Answer 2

Step-by-step explanation:

(1 - cos²A) sec²B + tan²B (1-sin²A) = sin²A + tan²B

• 1 - cos²A = sin²A

sin²A sec²B + tan²B - tan²A sin²A = sin²A + tan²B

sin²A ( sec²B - tan²A ) + tan²B = sin²A + tan²B

• sec²B - tan²A = 1

sin²A + tan²B = sin²A + tan²B