1-cos A -1-cos A-4-4 root 3-Find value of 1-sin-1-sin

1+cos A/1 cos A=4+4√(3) ⇒1+cos A/1 cos A+1=7+4√(3)+1 ⇒1+cos A+1 cos A/1 cos A=8+4√(3) ⇒2/1 cos A=2(4 +2√(2)) ⇒ 1 cos A =1/4+2√(3)×(4 2√(3))/4 2√(3) ⇒ 1 cos A = ...

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Byju's Answer

Standard X

Mathematics

Introduction to Trigonometry

1+cos A /1-co...

Question

(1+ cos A)/(1- cos A) = 4 + 4 root 3 =
Find value of $\frac{1 + s i n}{1 - s i n}$

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Solution

$\frac{1 + c o s A}{1 - c o s A} = 4 + 4 \sqrt{3} \Rightarrow \frac{1 + c o s A}{1 - c o s A} + 1 = 7 + 4 \sqrt{3} + 1 \Rightarrow \frac{1 + c o s A + 1 - c o s A}{1 - c o s A} = 8 + 4 \sqrt{3} \Rightarrow \frac{2}{1 - c o s A} = 2 (4 + 2 \sqrt{2}) \Rightarrow 1 - c o s A = \frac{1}{4 + 2 \sqrt{3}} \times \frac{(4 - 2 \sqrt{3})}{4 - 2 \sqrt{3}} \Rightarrow 1 - c o s A = \frac{4 - 2 \sqrt{3}}{4} = \frac{2 - \sqrt{3}}{2}$
$\Rightarrow 1 - \frac{2 - \sqrt{3}}{2} = c o s A \Rightarrow \frac{\sqrt{3}}{2} = c o s A$

$∴ \frac{1 + s i n A}{1 - s i n A} = \frac{1 + \frac{1}{2}}{1 - \frac{1}{2}} = \frac{\frac{3}{2}}{\frac{1}{2}} = 3 A n s$

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Similar questions

\frac{c o s A - s i n A + 1}{c o s A + s i n - 1} = c o s e c A + c o t A,

, using the identity

c o s e c^{2} A = 1 + c o t^{2} A

Q. If

C o s A + {C o s}^{2} A = 1

, then the value of

{S i n}^{2} A + {S i n}^{4} A

is :

Q. Prove that
1+cotA/1-secA=sin^2A/1-cosA

Q. If

\frac{cos A}{1 - sin A} + \frac{cos A}{1 + sin A} = 4

then find the value of

A

. (Here

0^{o} \leq A \leq 90^{o})

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(1+ cos A)/(1- cos A) = 4 + 4 root 3 = Find value of 1+sin1−sin

1+cosA1−cosA=4+4√3⇒1+cosA1−cosA+1=7+4√3+1⇒1+cosA+1−cosA1−cosA=8+4√3⇒21−cosA=2(4+2√2)⇒1−cosA=14+2√3×(4−2√3)4−2√3⇒1−cosA=4−2√34=2−√32 ⇒1−2−√32=cosA⇒√32=cosA ∴1+sinA1−sinA=1+121−12=3212=3Ans

(1+ cos A)/(1- cos A) = 4 + 4 root 3 =
Find value of $\frac{1 + s i n}{1 - s i n}$