Question

simplify 1-cos²theta/sin²( 90-theta)



pls slove it's hard for !me ​

Answer 1

Answer:

$\displaystyle{\boxed{\red{\sf\:\dfrac{1\:-\:\cos^2\:\theta}{\sin^2\:(\:90\:-\:\theta\:)}\:=\:\tan^2\:\theta}}}$

Step-by-step-explanation:

We have to simplify a trigonometric expression.

The given trigonometric expression is

$\displaystyle{\sf\:\dfrac{1\:-\:\cos^2\:\theta}{\sin^2\:(\:90\:-\:\theta\:)}}$

Now, we know that,

$\displaystyle{\pink{\sf\:\sin^2\:\theta\:+\:\cos^2\:\theta\:=\:1}}$

$\displaystyle{\implies\sf\:1\:-\:\cos^2\:\theta\:=\:\sin^2\:\theta}$

Now,

$\displaystyle{\pink{\sf\:\sin\:(\:90\:-\:\theta\:)\:=\:\cos\:\theta}}$

$\displaystyle{\implies\sf\:\sin^2\:(\:90\:-\:\theta\:)\:=\:\cos^2\:\theta}$

Now, by substituting these values in the given expression, we get,

$\displaystyle{\sf\:\dfrac{1\:-\:\cos^2\:\theta}{\sin^2\:(\:90\:-\:\theta\:)}}$

$\displaystyle{\implies\sf\:\dfrac{\sin^2\:\theta}{\cos^2\:\theta}}$

$\displaystyle{\implies\sf\:\left(\:\dfrac{\sin\:\theta}{\cos\:\theta}\:\right)^2}$

$\displaystyle{\implies\sf\:(\:\tan\:\theta\:)^2}$

$\displaystyle{\implies\sf\:\tan^2\:\theta}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:\dfrac{1\:-\:\cos^2\:\theta}{\sin^2\:(\:90\:-\:\theta\:)}\:=\:\tan^2\:\theta}}}}$

Answer 2

Answer:

$\sf \green {\boxed{ \bf \: \: \frac{1 - {cos}^{2} \theta }{ {sin}^{2} (90 - \theta) } = {tan}^{2} \theta \: \: }} \:$

Given,

$\sf \: \frac{1 - {cos}^{2} \theta }{ {sin}^{2}(90 - \theta)}$

We have to simplify this.

But, at first we have to know somethngs ! :

$\pink{ \sf{sin}^{2} \theta } + {cos}^{2} \theta = 1 \\ \pink{\sf \therefore \: {sin}^{2} \theta }= 1 - {cos}^{2} \theta$

And

$\odot \: \orange { \sf \: sin(90 - \theta)} = \blue{cos \theta }\\ \\ \sf\odot \: tan \theta = \frac{sin \theta}{cos \theta} \\ \\ \odot \: \sf \: ({sin \theta})^{2} = {sin}^{2} \theta \:$

Solution :

$\sf \: \frac{1 - {cos}^{2} \theta}{ {sin}^{2}(90 - \theta) } \\ \\ = \sf \: \frac{ \pink{ {sin}^{2} \theta} }{ { \{ \orange{sin(90 - \theta)}} \}^{2} } \\ \\ \sf = \frac{ \pink{{sin}^{2} \theta }}{ {( \blue{cos \theta}})^{2} } \\ \\ \sf = \frac{ \pink{ {sin}^{2} \theta}}{ \blue {{cos}^{2} \theta} } \\ \\ = \sf \purple{ {tan}^{2} \theta}$

So,

$\sf \green {\boxed{ \bf \: \: \frac{1 - {cos}^{2} \theta }{ {sin}^{2} (90 - \theta) } = {tan}^{2} \theta \: \: }}$