Question

plz solve this inverse trigonometry

Answer 1

canvert all of them in tan inverse
we get

tan inverse(5/12)+tan inverse(3/4) =tan inverse(56/33)

(by triangle low of trignomatery )

then we apply

L.H.L side


tan inverse(a+b) formula and

we get L.H.L =R.H.L

Answer 2

$\red\bullet\underline{\underline{\red{\bf Question}}}:$

To Prove :

• $\cos^{-1}\dfrac{12}{13}+\sin^{-1}\dfrac{3}{5}=\sin^{-1}\dfrac{56}{65}$

Concept to understand :

• First of all we will convert all terms to $\sin^{-1}$ . Then we will use formula , sum of inverse sin. Then we will get our answer.

$\red\bullet\underline{\underline{\red{\bf Solution}}}:$

$\cos^{-1}\dfrac{12}{13}+\sin^{-1}\dfrac{3}{5}=\sin^{-1}\dfrac{56}{65}$

$\large\green{\underbrace{\red{\rm L.H.S}}} :$

So,

$\sin^{-1} \dfrac{5}{13} +\sin^{-1} \dfrac{3}{5}$ $\quad \green{\{\bf Explanation\: is \: at \: bottom \: of\: the \: answer\}}$

$\red\bigstar\boxed{\sin^{-1}x+\sin^{-1}y = \sin^{-1}(y\sqrt{1-x^2}+x\sqrt{1-y^2}}$

$= \sin^{-1}\left( \dfrac{3}{5} \sqrt{1-\left( \frac{5}{13} \right)^2}+ \dfrac{5}{13} \sqrt{1-\left( \frac{3}{5} \right)^2}\right)$

$= \sin^{-1}\left( \dfrac{3}{5} \sqrt{1- \dfrac{25}{169} }+ \dfrac{5}{13} \sqrt{1-\dfrac{9}{25}}\right)$

$= \sin^{-1}\left( \dfrac{3}{5} \sqrt{ \dfrac{144}{169} }+ \dfrac{5}{13} \sqrt{\dfrac{16}{25}}\right)$

$= \sin^{-1}\left( \dfrac{3}{5} \times\dfrac{12}{13} + \dfrac{\cancel{5}}{13} \times\dfrac{4}{\cancel{5}} \right)$

$= \sin^{-1}\left( \dfrac{36}{65}+\dfrac{4}{13}\right)$

$= \sin^{-1} \dfrac{36+20}{65}$

$= \sin^{-1}\dfrac{56}{65}$

L.H.S = R.H.S

$\quad\underline{\green{\mathfrak{Hence\: verified..\displaystyle !!\,}}}$

• Note : We can convert $\cos^{-1} \dfrac{12}{13}$ to $\sin^{-1} \dfrac{5}{13}$ using trigonometric method.

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$\sf Let\:x = \cos^{-1}\dfrac{12}{13}\\ \cos x = \dfrac{12}{13}$

We know,

$\blue{\cos \theta = \sf\dfrac{Base}{Hyp.}}$

So we have ,

• hyp = 13
• base = 12

We can find perpendicular using Pythagoras theorem,

$\sf H ^2= B^2+P^2 \\ \sf 13^2 = 12^2 + P \\\sf 169-144 = P \\\sf P^2 = 25 \\\sf\boxed{P=5}$

We also know ,

$\blue{\sin\theta = \sf\dfrac{perp.}{hyp.} }$

So,

$\sin x = \dfrac{5}{13}\\\sin^{-1}\dfrac{5}{13}$

Henceforth, we can use $\: \cos^{-1} \dfrac{12}{13} = \sin^{-1} \dfrac{5}{13}$

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