Question

In a ∆ABC,right angled at B and coses A =√2,the 1/tanA+sinA/1+cosA
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Answer 1

Answer:

[Do not consider theta=30° in figure, it is mistake]

Step-by-step explanation:

Given: cosec A=2.

Cosec A=1/sin A=AC/BC=2/1

AC=2, BC=1

By Pythagoras theorem:

AC²=AB²+BC2

=>2²-1²=AB²

=>4-1=AB²

=>√3=AB.

tanA=BC/AB=1/√3

sinA=BC/AC=1/2.

cosA=AB/AC=√3/2

1/tanA+sinA/1+cosA

=√3+2+√3

=2√3+2.

Hope helps u.

Answer 2

Answer:

√2 – 1

Step-by-step explanation:

As per the information provided in the question, We have :

• Angle B = 90°
• Cosec A = √2

In order to find 1/tanA+sinA/1+cosA, We need to find sides of the triangle. In order to do that we will use Pythagoras theorem.

Trigonometric ratio of Cosec is hyp/per. Thus, Hyp is √2 units and per is 1 unit.

Applying Pythagoras theorem, To find base,

⇒ AC² = AB² + BC²

⇒ (√2)² = AB² + 1²

⇒ 2 = AB² + 1

⇒ AB² = 2 - 1

⇒ AB² = 1

⇒ AB = √1 = 1

∴ Base of the triangle is 1 unit.

Finding 1/tanA+sinA/1+cosA :

Equation - 1/tanA+sinA/1+cosA

Where,

• Tan A = per/base = 1/1 = 1
• Sin A = per/hyp = 1/√2
• Cos A = base/hyp = 1/√2

Substituting the values.

$\longmapsto \rm \dfrac{1}{ \tan(A) } + \dfrac{ \sin(A) }{1 + \cos(A) }$

$\longmapsto \rm \dfrac{1}{ 1 } + \dfrac{ \sin(A) }{1 + \cos(A) }$

$\longmapsto \rm \dfrac{ \dfrac{1}{ \sqrt{2} } }{1 + \dfrac{1}{ \sqrt{2} } }$

$\longmapsto \rm \dfrac{ 1 }{ \sqrt{2} \times 1 + \dfrac{1}{ \sqrt{2} } }$

$\longmapsto \rm \dfrac{ 1 }{ \sqrt{2} \times \dfrac{ \sqrt{2} + 1}{ \sqrt{2} } }$

By cancelling √2,

$\longmapsto \rm \dfrac{ 1 }{ {\sqrt{2} + 1}{ } }$

On rationalising,

$\longmapsto \rm \sqrt{2} - 1$

∴ 1/tanA+sinA/1+cosA = √2 – 1.