Question

solve for A if,
$\frac{sin A}{1+cos A} +\frac{1+ cos A}{sin A} =4$

Answer 1

Question :---- Solve for A if sinA/(1+cosA) + (1+cosA)/sinA = 4 .

Formula used :---

• (a+b)² = a² + b² + 2ab
• sin²A + cos²A = 1
• 1/sinA = cosecA .
• cosec30° = 2 .

Solution :----

Solving the LHS , by taking the LCM first we get,

→ sinA/(1+cosA) + (1+cosA) / sinA

→ [sin²A + (1 + cosA)²] / sinA(1 + cosA)

using (a+b)² = a² + b² + 2ab in numerator now,

→ (sin²A + 1 + cos²A + 2cosA) / sinA(1 + cosA)

using sin²A + cos²A = 1 in numerator now,

→ ( 2 + 2cosA ) / sinA(1+cosA)

→ 2(1+cosA) / sinA(1+cosA)

→ 2/sinA

→ 2cosecA

Now, putting this to RHS we get,

→ 2cosecA = 4

Dividing both sides by 2 ,

→ cosecA = 2

Now putting 2 = cosec30° ,

→ cosecA = cosec30°

→ A = 30°

Hence, value of A is 30° .

Answer 2

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

$\leadsto \tt {\dfrac {sin A}{1+cosA}+\dfrac {1+ cos A}{sin A} =4}$

Taking LCM,

$\leadsto\tt {\dfrac {{sin}^{2}A+{(1+cosA)}^{2}}{sinA+sinA\:cosA}=4}$

We know that, $\tt {{sin}^{2}\theta+{cos}^{2}\theta=1}.$

Therefore,

$\leadsto\tt {\dfrac {1+1+2\:cosA}{sinA+sinA\:cosA)}=4}$

$\leadsto\tt {\dfrac {2+2\:cosA}{sinA+sinA\:cosA}=4}$

Taking common,

$\leadsto\tt {\dfrac {2 \cancel {(1+cosA)}}{sinA\cancel {(1+cosA)}}=4}$

$\leadsto\tt {\dfrac {2}{sinA}=4}$

$\leadsto\tt {2\:cosecA=2}$

$\leadsto \tt {cosecA=2}$

We know that $\leadsto \tt {cosec \:30^{\circ}=2.}$

Hence,

$\leadsto\tt {cosecA=cosec\:30^{\circ}}$

$\huge\leadsto {\boxed {\tt {So, A = 30^{\circ}}}}$

$\rule{200}2$