Question

cosQ/1 +sinQ + 1+sinQ/cosQ = 2secQ​

Answer 1

To Prove

$\sf\frac{cos\theta}{1 + sin\theta} + \frac{1 + sin\theta}{cos\theta}$ = $\sf 2sec\theta$

$\rule{100}2$

Proof

$\sf\frac{cos\theta}{1 + sin\theta} + \frac{1 + sin\theta}{cos\theta}$

$\sf\frac{cos^2\theta + (1 + sin\theta)^2}{cos\theta(1 + sin\theta)}$

(By making denominators equal)

$\sf\implies\frac{cos^2\theta + 1 + sin^2\theta + 2sin\theta}{cos\theta(1 + sin\theta)}$

(Using (a+ b)² = a² + b² + 2ab)

$\sf\implies\frac{1 + 1 + 2sin\theta}{cos\theta(1 + sin\theta)}$

(using, cos²∅ + sin²∅ = 1)

$\sf\implies\frac{2 + 2sin\theta}{cos\theta(1 + sin\theta)}$

$\sf\implies\frac{2(1 + sin\theta)}{cos\theta(1 + sin\theta)}$

$\sf\implies\frac{2}{cos\theta}$

(cancelling (1 + sin∅) from both numerator and denominator)

$\sf\implies 2sec\theta$

(since, $\sf\frac{1}{cos\theta} = sec\theta$)

Hence proved.

Answer 2

*I am using A as angle*

$\huge\underline\mathfrak\pink{Question-}$

Prove that -

$\sf\dfrac{CosA}{1+SinA}$ + $\sf\dfrac{1+SinA}{CosA}$ = $\sf{2SecA}$

$\huge\underline\mathfrak\pink{Solution-}$

$\sf\dfrac{CosA}{1+SinA}$ + $\sf\dfrac{1+SinA}{CosA}$ = $\sf{2SecA}$

LHS,

$\leadsto$ $\sf\dfrac{CosA}{1+SinA}$ + $\sf\dfrac{1+SinA}{CosA}$

By taking LCM,

$\leadsto$ $\sf\dfrac{ { \cos(A) }^{2} + {(1 + \sin(A)) }^{2} }{(1 + \sin(A))(cos(A)) }$

$\leadsto$ $\sf\dfrac{ { \cos(A) }^{2} + 1 + { \sin(A) }^{2} + 2 \sin(A) }{(1 + \sin(A)) \cos(A) }$

$\leadsto$ $\sf\dfrac{1+1+2SinA}{(1+SinA)(CosA)}$

$\leadsto$ $\sf\dfrac{2(1+SinA)}{(1+SinA)CosA}$

$\leadsto$ $\sf\dfrac{2(\cancel{1+SinA)}}{(\cancel{1+SinA)}CosA}$

$\leadsto$ $\sf\dfrac{2}{CosA}$

$\leadsto$ $\sf{2SecA}$

$\leadsto$ RHS

Hence proved!

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$\large\bold\blue{Identities\:used-}$

★ ( a + b )² = a² + b² + 2ab

★ Sin²A + Cos²A = 1

★ $\sf\dfrac{1}{CosA}$ = SecA