Question

can some one help with this sum on trignomentric ratios

Answer 1

Question :-

If

$\dfrac{ \sec \theta - \tan \theta}{ \sec \theta + \tan \theta} = \sf \: \dfrac{1}{4}$

then find sin θ.

Answer:-

Given:-

$\: \dfrac{ \sec \theta - \tan \theta}{ \sec \theta + \tan \theta} = \sf \: \dfrac{1}{4} \\ \\ \\ \implies \sf \: ( \sec \theta - \tan \theta) \times \frac{1}{ \sec \theta + \tan \theta } = \frac{1}{4}$

we know that,

sec² θ - tan² θ = 1

• a² - b² = (a + b)(a - b)

So,

⟹ (sec θ + tan θ) (sec θ - tan θ) = 1

⟹ sec θ - tan θ = 1/ sec θ + tan θ -- equation (1)

Hence,

⟹ (sec θ - tan θ) (sec θ - tan θ) = 1/4

⟹ (sec θ - tan θ)² = (1/4)

using sec θ = 1/cos θ and tan θ = sin θ / cos θ we get,

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

using the identity cos² θ = 1 - sin² θ we get,

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

Again using a² - b² = (a + b)(a - b) we get,

$\: \implies \sf \: \frac{(1 - \sin \theta) \cancel{(1 - \sin \theta)} }{(1 + \sin \theta) \cancel{(1 - \sin \theta)}} = \frac{1}{4} \\ \\ \\ \implies \sf \: \frac{1 - \sin \theta }{1 + \sin \theta} = \frac{1}{4} \\ \\ \\ \implies \sf \:4(1 - \sin \theta) = 1 + \sin \theta \\ \\ \\ \implies \sf \:4 - 4 \sin \theta = 1 + \sin \theta \\ \\ \\ \implies \sf \:4 - 1 = \sin \theta + 4 \sin \theta \\ \\ \\ \implies \sf \:3 = 5 \sin \theta \\ \\ \\ \implies \boxed{ \red{\sf \: \sin \theta = \frac{3}{5}}}$

Answer 2

$\begin{gathered}\begin{gathered}\bf \: Given \:  - \begin{cases} &\sf{\dfrac{ \sec \theta - \tan \theta}{ \sec \theta + \tan \theta} = \sf \: \dfrac{1}{4}} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \:To\:find\:- \begin{cases} &\sf{sin\theta}  \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\Large{\bold{\blue{\underline{Formula \: Used \::}}}} \end{gathered}$

$\rm :\implies\: \boxed{ \pink{ \bf \:tan\theta \: = \tt \:\dfrac{sin\theta}{cos\theta} }}$

$\rm :\implies\: \boxed{ \pink{ \bf \: sec\theta \: = \tt \: \dfrac{1}{cos\theta} }}$

$\large\underline\purple{\bold{Solution :- }}$

$\rm :\implies\:\dfrac{ \sec \theta - \tan \theta}{ \sec \theta + \tan \theta} = \sf \: \dfrac{1}{4}$

$\rm :\implies\:4sec\theta - 4tan\theta = sec\theta + tan\theta$

$\rm :\implies\:4sec\theta - sec\theta = tan\theta + 4tan\theta$

$\rm :\implies\:3 \: sec\theta \: = \: 5 \: tan\theta$

$\rm :\implies\:\dfrac{3}{cos\theta} = \: \dfrac{5 \: sin\theta}{cos\theta}$

$\rm :\implies\:3 \: = \: 5 \: sin\theta$

$\rm :\implies\: \boxed{ \pink{ \bf \:sin\theta \: = \tt \: \dfrac{3}{5} }}$

Additional Information:-

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

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$