Question

$The\: value\: of\: \dfrac{cos \theta }{1+sin \theta} \: is\: equal\: to$

a) $tan \bigg( \dfrac{\theta}{2} - \dfrac{\pi}{4} \bigg)$

b) $tan \bigg( \dfrac{-\pi}{4} - \dfrac{\theta}{2} \bigg)$

c) $tan \bigg( \dfrac{\pi}{4} - \dfrac{\theta}{2} \bigg)$

d) $tan \bigg( \dfrac{ \pi}{4} + \dfrac{\theta}{2} \bigg)$​

Answer 1

Question :-- Find the value of cosA /(1+sinA) ?

Formula used :-

• CosA = (1-tan²A/2) / (1 + tan²A/2)
• SinA = 2tanA/2 / ( 1+ tan²A/2)
• a² + b² + 2ab = (a+b)²
• a² - b² = (a+b)(a-b)
• 1 = tan45° = tanπ/4

Solution :-

→ cosA/(1+sinA)

Putting values told above , we get,

→ {(1-tan²A/2) / (1 + tan²A/2)} / 1 + {2tanA/2 / ( 1+ tan²A/2)}

Now, Taking LCM of Denominator we get,

→ {(1-tan²A/2) / (1 + tan²A/2)} / [ ( 1 + 2tanA/2 + tan²A/2 ) / ( 1+ tan²A/2) ]

Now, as we can see that, both denominator part have ( 1+ tan²A/2) , so both will be cancel ,

we get,

→ [ 1 - tan²A/2 ] / [ 1 + 2tanA/2 + tan²A/2 ]

Now, Denominator, using a² + 2ab + b² = (a+b)² we get,

→ [ 1 - tan²A/2 ] / [ 1 + tanA/2]²

→ Now, using (a² - b² = (a+b)(a-b) in Numerator we get,

→ ( 1 + tanA/2) ( 1 - tanA/2) / (1+tanA/2)²

(1+tanA/2) will be cancel from both sides ,

→ ( 1 - tanA/2) / (1 + tanA/2) ----------- Equation (1)

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Now,

we know that, Tan 45° = tanπ/4 = 1 .

Lets Try to Expand , Tan(π/4 - A) by using the Formula

☛ Tan(A - B) = (TanA - TanB) / (1 + TanA*TanB)

Expanding Tan(π/4 - A/2) , with this we get,

➺ Tan(π/4 - A/2) = ( Tanπ/4 - TanA/2) / ( 1 + Tanπ/4 * TanA/2) .

Now, putting Tanπ/4 as 1 , we get,

➺ Tan(π/4 - A/2) = ( 1 - tanA/2) / (1 + tanA/2) = Equation (1) . (✪✪ Hence Proved ✪✪)

So, we can say that, cosA/(1+sinA) is Equal to Tan(π/4 - A/2).

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❦❦ Remember :- if the examiner asked Value of cosA / (1 - sinA) , Than, it will be Equal to Tan(π/4 + A/2).

Answer 2

Solution...in the attachment.