Question

If sin θ = a/b, Then find all Trigonometric Ratios​

Answer 1

Answer:

hope it helps you my friend

Answer 2

Answer:

We know that :

$\sf{\sin\theta = Opposite\:Side/Hypotenuse}$

$\longrightarrow\sf{a/b = Opposite\:Side/Hypotenuse}$

$\bigstar$ From this Information ;

$\longrightarrow$ Opposite Side = a $\&$ Hypotenuse = b

We can apply Pythagoras Theorem (Pythagoras Theorem states that Square of Hypotenuse is equal to Square of Other two Sides) in The triangle and find the measure of Adjacent Side

$\boxed{\rm{(Hypotenuse)^2 = (Opposite\: Side)^2 + (Adjacent\: Side)^2}}$

$\longrightarrow\rm{b^2 = a^2 + (Adjacent\: Side)^2\quad...\:\sf{Since\:Opposite\:Side = a\:\&\: Hypotenuse = b }}$

$\longrightarrow\rm{(Adjacent\: Side)^2 = b^2 - a^2\quad...\:\sf{Subtracting\:a^2\:from\:both\:Sides\:of\:Equation}}$

$\longrightarrow\rm{Adjacent\: Side = \sqrt{b^2 - a^2}\quad...\:\sf{Taking\:Square\:Root\:on\:both\:Sides\:of\:Equation}}$

$\bigstar$ Now, From the Calculations ;

$\longrightarrow$ Opposite Side = a, Adjacent Side = $\sqrt{\sf{b^2-a^2}}$ $\&$ Hypotenuse = b

$\boxed{\begin{aligned}&\bullet\:\sin\theta = \rm{Opposite\:Side/Hypotenuse = a/b}\\&\bullet\:\cos\theta = \rm{Adjacent\:Side/Hypotenuse = \sqrt{b^2-a^2}/b}\\&\bullet\:\tan\theta = \rm{Opposite\:Side/Adjacent\:Side = a/\sqrt{b^2-a^2}}\\&\bullet\:\csc\theta = \rm{Hypotenuse/Opposite\:Side = b/a}\\&\bullet\:\sec\theta = \rm{Hypotenuse/Adjacent\:Side = b/\sqrt{b^2-a^2}}\\&\bullet\:\cot\theta = \rm{Adjacent\:Side/Opposite\:Side = \sqrt{b^2-a^2}/a}\\\end{aligned}}$

$\large{----------\textbf{ Alternate Method }----------}$

• We know that :

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

$\longrightarrow\rm{(a/b)^2 + \cos^2\theta = 1\longrightarrow\rm{a^2/b^2 + \cos^2\theta = 1}\quad...\:\sf{Since\:\sin\theta\:is\:equal\:to\:a/b}}$

$\longrightarrow\rm{\cos^2\theta = 1 - a^2/b^2\quad...\:\sf{Subtracting\:a^2/b^2\:from\:both\:Sides\:of\:the\:Equation}}$

$\longrightarrow\rm{\cos^2\theta = b^2/b^2 - a^2/b^2 = (b^2-a^2)/b^2\quad...\:\sf{Taking\:LCM\:of\:1\:and\:b^2}}$

$\longrightarrow\rm{\cos\theta = \sqrt{(b^2-a^2)/b^2} = \sqrt{b^2-a^2}/b\quad...\:\sf{Taking\:Square\:Root\:on\:both\:sides}}$

• We know that :

$\sf{\tan\theta = \sin\theta/\cos\theta}$

$\longrightarrow\rm{\tan\theta=a/b\div \sqrt{b^2-a^2}/b\quad...\:\sf{Since\:\sin\theta = a/b\:and\:\cos\theta = \sqrt{b^2-a^2}/b}}$

$\longrightarrow\rm{\tan\theta=a/b\times b/\sqrt{b^2-a^2} = ab/b\sqrt{b^2-a^2} = a/\sqrt{b^2-a^2}}$

• We know that :

$\sf{\csc\theta = 1/\sin\theta}$

$\longrightarrow\rm{\csc\theta = 1\div a/b = 1\times b/a = b/a\quad...\:\sf{Since\:\sin\theta\:is\:equal\:to\:a/b}}$

We know that : $\sf{\sec\theta = 1/\cos\theta}$

$\longrightarrow\rm{\sec\theta = 1\div \sqrt{b^2-a^2}/b} = b/\sqrt{b^2-a^2}\quad...\:\sf{Since\:\cos\theta\:is\:equal\:to\:\sqrt{b^2-a^2}/b}$

• We know that :

$\sf{\cot\theta = 1/\tan\theta}$

$\longrightarrow\rm{\cot\theta = 1\div a/\sqrt{b^2-a^2} = \sqrt{b^2-a^2}/a\quad...\:\sf{Since\:\tan\theta\:is\:equal\:to\:a/\sqrt{b^2-a^2}}}$

$\bigstar$ From the Calculations ;.

$\boxed{\begin{aligned}&\bullet\:\sin\theta = \rm{Opposite\:Side/Hypotenuse = a/b}\\&\bullet\:\cos\theta = \rm{Adjacent\:Side/Hypotenuse = \sqrt{b^2-a^2}/b}\\&\bullet\:\tan\theta = \rm{Opposite\:Side/Adjacent\:Side = a/\sqrt{b^2-a^2}}\\&\bullet\:\csc\theta = \rm{Hypotenuse/Opposite\:Side = b/a}\\&\bullet\:\sec\theta = \rm{Hypotenuse/Adjacent\:Side = b/\sqrt{b^2-a^2}}\\&\bullet\:\cot\theta = \rm{Adjacent\:Side/Opposite\:Side = \sqrt{b^2-a^2}/a}\\\end{aligned}}$