Question

if sin theta is equal to p by q then the value of tan theta + secant theta​

Answer 1

This from a fellow friend

Answer 2

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:sin \theta \: = \: \dfrac{p}{q}$

We know,

$\rm :\longmapsto\:sin \theta \: = \: \dfrac{opposite \: side}{hypotenuse}$

$\rm :\longmapsto\: \dfrac{opposite \: side}{hypotenuse} = \dfrac{p}{q}$

We know, By pythagoras theorem

$\rm :\longmapsto\: {hypotenuse}^{2} = {opposite \: side}^{2} + {adjacent \: side}^{2}$

$\rm :\longmapsto\: {q}^{2} = {p}^{2} + {adjacent \: side}^{2}$

$\rm :\longmapsto\: {q}^{2} - {p}^{2} = {adjacent \: side}^{2}$

$\rm :\longmapsto\: {adjacent \: side} = \sqrt{ {q}^{2} - {p}^{2}}$

Now,

$\purple{\rm :\longmapsto\:tan\theta = \dfrac{opposite \: side}{adjacent \: side}}$

$\rm\implies \:tan\theta = \dfrac{p}{\sqrt{ {q}^{2} - {p}^{2} } }$

Now,

$\purple{\rm :\longmapsto\:sec\theta = \dfrac{hypotenuse}{adjacent \: side}}$

$\rm\implies \:sec\theta \: = \: \dfrac{q}{\sqrt{ {q}^{2} - {p}^{2} } }$

Now, Consider

$\purple{\rm :\longmapsto\:tan\theta + sec\theta }$

$\rm \:  =  \: \dfrac{p}{\sqrt{ {q}^{2} - {p}^{2} } } + \dfrac{q}{\sqrt{ {q}^{2} - {p}^{2} } }$

$\rm \:  =  \: \dfrac{p + q}{\sqrt{ {q}^{2} - {p}^{2} } }$

$\rm \:  =  \: \dfrac{p + q}{\sqrt{(q + p)(q - p)} }$

$\rm \:  =  \: \sqrt{\dfrac{q + p}{q - p} }$

Hence,

$\\ \purple{\rm\implies \:\boxed{\tt{ \rm \: tan\theta + sec\theta =  \: \sqrt{\dfrac{q + p}{q - p}}}}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1