Question

If sin theta=a/b then prove that sec theta+ tan theta = √b+a/√b-a

Answer 1

EXPLANATION:

In detail solution.

$let \: theta = \alpha \\ \sin\alpha = \frac{a}{b} = \frac{p}{h} \\ b = \sqrt{ {b}^{2} - {a}^{2} } \\ now \\ \sec\alpha + \tan\alpha = \frac{h}{b} + \frac{p}{b} \\ \: \: \: \: \: \: \: \: \: \: \: = \frac{h + p}{b} \\ \: \: \: \: = \frac{b + a}{ \sqrt{ {b}^{2} - {a}^{2} } } \\ = \: \: \frac{ \sqrt{ {(b + a)}^{2} } }{ \sqrt{ {b}^{2} - {a}^{2} } } \\ = \sqrt{ \frac{b + a}{b - a} }$

Answer 2

Topic:-

Trigonometry

Question:-

$if \: sin \theta = \frac{a}{b} then \: prove \: sec \theta + tan \theta \: = \frac{ \sqrt{b + a} }{ \sqrt{b - a} }$

Answer:-

First Taking LHS

$sec \theta + tan \theta$

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

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

$= \frac{1 + sin \theta}{ \sqrt{1 - {sin}^{2}theta } }$

$we \: know \: that \: \\ \\ sin \theta = \frac{a}{b}$

$= \frac{1 + \frac{a}{b} }{ \sqrt{1 - { (\frac{a}{b} )}^{2} } }$

$= \frac{ (\frac{b + a}{b}) }{ \sqrt{ \frac{ {b}^{2} - {a}^{2} }{ {b}^{2} } } }$

$= \frac{ \frac{b + a}{b} }{ \frac{ \sqrt{ {b}^{2} - {a}^{2} } }{ \sqrt{ ({b}^{2}) } } }$

$= \frac{ \frac{b + a}{b} }{ \frac{ \sqrt{b + a} \times \sqrt{b - a} }{b} }$

$= \frac{b + a}{ \sqrt{b + a} \times \sqrt{b - a} }$

$= \frac{ \sqrt{b + a} }{ \sqrt{b - a} }$

$= \sqrt{ \frac{b + a}{b - a} }$

= RHS

Hence Proved//

More Information:-

Trigon metric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

Trigometric relations

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

Trigonmetric ratios

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj