Question

tan theta/tan(90°-theta) + sin(90°-theta)/cos theta ​

Answer 1

Answer:

sec²θ

Step-by-step explanation:

tan(90-θ) = cotθ  and sin(90-θ) = cosθ

∴ tanθ/tan(90-θ) + sin(90-θ)/cosθ

= tanθ/cotθ + cosθ/cosθ

= tanθ÷ 1/tanθ + 1

= tan²θ +1

= sec²θ

Answer 2

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Find :

$\frac{ \tan \theta}{ \tan(90 - \theta) } \: + \: \frac{ \sin(90 - \theta) }{ \cos\theta }$

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$\frac{ \tan \theta}{ \tan(90 - \theta) } \: + \: \frac{ \sin(90 - \theta) }{ \cos\theta }$

↬ $\frac{ \tan \theta}{ \tan(1 \times 90 - \theta) } \: + \: \frac{ \sin(1 \times 90 - \theta) }{ \cos\theta }$

↬ $\frac{ \tan \theta}{ \cot\theta}\: + \frac{ \cos\theta }{ \cos\theta }$

↬ $(\frac{ \sin\theta }{ \cos\theta } \div \frac{ \cos\theta }{ \sin\theta }) + 1$

↬ $\frac{ \sin\theta }{ \cos\theta } \times \frac{ \sin\theta }{ \cos\theta } + 1$

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

↬ $\frac{ { \sin }^{2}\theta }{ { \cos }^{2} \theta} + 1$

↬ $\frac{ {sin}^{2} \theta \: + \: {cos}^{2}\theta }{ {cos}^{2}\theta }$

↬ $\frac{1}{ {cos}^{2}\theta }$

↬ ${sec}^{2} \theta$

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$sin \: \theta = \frac{p}{h} \\ \\ cos \: \theta = \frac{b}{h} \\ \\ tan \: \theta = \frac{p}{b}$

$cosec \: \theta = \frac{h}{p} = \frac{1}{sin \: \theta}$

$sec \: \theta = \frac{h}{b} = \frac{1}{cos \: \theta }$

$cot \: \theta = \frac{b}{p} = \frac{1}{tan \: \theta }$