Question

If cos ( A + B ) = 0 and cot ( A – B ) = √3, find the value of:

a) sec A tan B – cot A sin B

b) cosec A cot B + sin A tan B​

Answer 1

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Step-by-step explanation:

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Answer 2

Solution :-

Given ,

• cos( A + B ) = 0
• cot( A - B ) = √3

We need to find ,

• secA tanB - cotA sinB
• cosecA cotB - sinA tanB

If ,

• cos( A + B ) = 0

Substituting 0 = cos90°

→ cos( A + B ) = cos90°

→ A + B = 90°

→ A = 90° - B …eq(1)

• cot( A - B ) = √3

Substituting √3 = cot30°

→ cot( A - B ) = cot30°

→ A - B = 30°

→ A = 30° + B …eq(2)

Now , from equation 1 & 2

• 30° + B = 90° - B

• 90° - 30° = B + B

• 60° = 2B

• B = 60°/2

• B = 30°

Now , finding angle A using eq(1)

• A = 90° - B

Substituting value of angle B

• A = 90° - 30°

• A = 60°

Now , coming to the question

◉ secA tanB - cotA sinB

Substituting the values of angles A & B

⇒ sec60° tan30° - cot60° sin30°

⇒ 2(1/√3) - (1/√3)(1/2)

⇒ 2/√3 - 1/2√3

⇒ 4 - 1/2√3

⇒ 3/2√3

By Rationalizing the denominator

⇒ 3(2√3)/(2√3)(2√3)

⇒ 6√3/4(3)

⇒ 6√3/12

⇒ secA tanB - cotA sinB = √3/2

◉ cosecA cotB + sinA tanB

Substituting the values of angles A & B

⇒ cosec60° cot30° + sin60° tan30°

⇒ 2/√3 ( √3 ) + √3/2 ( 1/√3)

⇒ 2 + 1/2

⇒ ( 4 + 1/2)

⇒ cosecA cotB + sinA tanB = 5/2