Question

if cos ( A + B ) = 0 and cot ( A-B ) = √3 find the value of secA.tanB - cotA . sinB

Answer 1

Given :  Cos ( A + B ) = 0 and Cot ( A - B ) = √3.

We Know that :

Cos90° = 0

∵ Cos 90° = Cos (A + B)

∴ (A+B)= 90            (i)

Cot30° = √3

∵ Cot 30° = Cot (A - B)

∴ (A-B)= 30            (ii)

Adding Equations (i)&(ii) ;

A+B = 90

A-B= 30

2A = 120°

A = 60° And B = 30°

Now,

secA.tanB - cotA . sinB

= sec60.tan30 - cot60 . sin30

= $2\times\frac{1}{\sqrt{3}} -\frac{1}{\sqrt{√3}}\times\frac{1}{2}}$

=$\bold{\frac{2}{\sqrt{3}} - \frac{1}{2\sqrt{3}}}$

= $\bold{\frac{4-1}{2\sqrt{3}}}$        

=$\bold{\frac{\sqrt{3}}{2}}$

Answer 2

Answer:

√3/2

Step-by-step explanation:

(i)

⇒ cos(A + B) = 0

⇒ cos(A + B) = cos 90°

⇒ A + B = 90°

(ii)

⇒ Cot(A - B) = √3

⇒ Cot(A - B) = cot 30°

⇒ A - B = 30°

On adding (i) & (ii).

A + B = 90

A - B = 30

---------------

2A = 120

A = 60°

Place A = 60° in (i).

⇒ A + B = 90

⇒ 60 + B = 90

⇒ B = 30.

SecA. tanB - cotA.sinB

⇒ sec(60) * tan(30) - cot(60) * sin(30)

⇒ 2 * (1/√3) - (1/√3) * (1/2)

⇒ (2/√3) - (1/2√3)

⇒ (4 - 1)/(2√3)

⇒ 3/2√3

⇒ (√3 * √3)/2√3

⇒ √3/2.

Hope it helps you.

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