Question

solve cos theta divided by 1 - sin theta + cos theta divided by 1 + sin theta is equals to 4​

Answer 1

Answer :-

θ = 60°

Explanation :-

$\rm \dfrac{cos \theta}{1 - sin \theta } + \dfrac{cos \theta}{1 + sin \theta} = 4$

$\text{Taking LCM}$

$\rm \implies \dfrac{cos \theta(1 + sin \theta) + cos \theta(1 - sin \theta)}{(1 - sin \theta)(1 + sin \theta) }= 4$

$\rm \implies \dfrac{cos \theta + cos \theta .sin \theta+ cos \theta - cos \theta .sin \theta}{1^{2} - sin^{2} \theta }= 4$

$\boxed{\boxed{ \bf \because (x - y)(x + y) = {x}^{2} - {y}^{2} }}$

$\rm \implies \dfrac{2cos \theta }{1 - sin^{2} \theta }= 4$

$\rm \implies \dfrac{2cos \theta }{cos^{2} \theta}= 4$

$\boxed{\boxed{ \bf \because 1 - sin^{2} \theta = cos^{2} \theta }}$

$\rm \implies \dfrac{2cos \theta }{cos\theta.cos \theta}= 4$

$\rm \implies \dfrac{2 }{cos \theta}= 4$

$\rm \implies 2 = 4cos \theta$

$\rm \implies \dfrac{2}{4} = cos \theta$

$\rm \implies \dfrac{1}{2} = cos \theta$

$\rm \implies cos \ 60^{ \circ} = cos \theta$

$\boxed{\boxed{ \bf \because cos \ 60^{ \circ} = \dfrac{1}{2} }}$

$\rm \implies 60^{ \circ} = \theta$

$\rm \implies \theta = 60^{ \circ}$

$\huge{\boxed{ \boxed{\tt{ \therefore \theta = 60^{ \circ}}}}}$

Answer 2

Given:

$\\ \sf \: \dfrac{cos \theta }{1 - sin \: \theta} + \dfrac{cos \: \theta}{1 + sin \: \theta } = 4$

To find:

$\sf value \: of \: \theta$

Explanation:

$\implies \: \sf \: \dfrac{cos \: \theta(1 + sin \theta) +cos \: \theta(1 - sin \theta) }{(1 + sin \:\theta)(1 - sin \:\theta)} = 4 \\ \\ \implies \: \sf \: \dfrac{cos\theta + cos\theta \: sin\theta + cos \theta - cos\theta \: sin\theta}{1 - {sin}^{2}\theta \: } = 4$

we know that

$\underline{\tt (a+b)(a-b)=a^2-b^2}$

$\underline{\tt 1-sin^2 \theta=cos^2 \theta}$

$\implies \: \sf \: \dfrac{2cos\theta}{ {cos}^{2} \theta} = 4 \\ \\ \implies \: \sf \dfrac{2}{cos \theta} = 4 \\ \\ \implies \: \sf cos \theta = \dfrac{1}{2}$

$\huge{\boxed{\boxed{\tt \theta=60\degree}}}$

More information:

↔Cosec²a-cot²a=1

↔sec²a-tan²a=1

↔sin²a+cos²a=1