
Solve the following: $\tan (1/2{{\sin }^{-1}}3/4)$
Answer
430.2k+ views
Hint: Revise all the formulas of trigonometry and all the properties of inverse trigonometric functions. The inverse trigonometric functions are the inverse functions of the trigonometric functions for example inverse of sine, cosine, tangent etc.
Complete step by step solution:
We have to solve $\tan (1/2{{\sin }^{-1}}3/4)$ for that let us assume that
$1/2{{\sin }^{-1}}3/4=\theta $
Now by cross multiplication
$1/2{{\sin }^{-1}}3/4=\theta $ becomes
${{\sin }^{-1}}3/4=2\theta $ ---- (1)
Now by multiplying sin on both sides of the equation(1)
$\sin ({{\sin }^{-1}}3/4)=\sin 2\theta $
$\sin 2\theta =3/4$ ---- (2)
Now by using the formula $\sin 2\theta =2\tan \theta /(1+{{\tan }^{2}}\theta )$ on equation (2)
$2\tan \theta /(1+{{\tan }^{2}}\theta )=3/4$
Now by cross multiplication
$2\tan \theta /(1+{{\tan }^{2}}\theta )=3/4$ becomes
$4(2\tan \theta )=3(1+{{\tan }^{2}}\theta )$ ----- (3)
By solving the brackets of equation (3)
$8\tan \theta =3+3{{\tan }^{2}}\theta $
$3{{\tan }^{2}}\theta -8\tan \theta +3=0$ ------- (4)
To find the roots of a quadratic equation we use the formula
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
So by using the above formula on equation (4)
$\tan \theta =-(-8)\pm \sqrt{64-(4 \times 3 \times 3}/2 \times 3$
So, $\tan \theta =4\pm \sqrt{7}/3$ ----- (5)
Now by taking tan inverse on both sides of equation (5)
$\theta ={{\tan }^{-1}}\left[ 4\pm \sqrt{7}/3 \right]$
$\tan \theta =4\pm \sqrt{7}/3$
As $\theta =1/2{{\sin }^{-1}}3/4$
So, $\tan \theta =4\pm \sqrt{7}/3$
Since ,
$\begin{align}
& -\pi /2\le {{\sin }^{-1}}3/4\le \pi /2 \\
& -\pi /4\le 1/2{{\sin }^{-1}}3/4\le \pi /4 \\
\end{align}$
Therefore, $\tan (-\pi /4)\le \tan 1/2({{\sin }^{-1}}3/4)\le \tan \pi /4$
$-1\le \tan (1/2{{\sin }^{-1}}3/4)\le 1$
Since, $4+\sqrt{7}/3>1$ so it is ignored
Therefore, $\tan (1/2{{\sin }^{-1}}3/4)=4-\sqrt{7}/3$.
Note:
There is a restriction on $\sin \theta $ i.e. $-\pi /2\le {{\sin }^{-1}}\theta \le \pi /2$. So all the values which are greater than one should be ignored. Always use the correct trigonometric formula to solve a particular equation as using the wrong formula leads towards the wrong answer.
Complete step by step solution:
We have to solve $\tan (1/2{{\sin }^{-1}}3/4)$ for that let us assume that
$1/2{{\sin }^{-1}}3/4=\theta $
Now by cross multiplication
$1/2{{\sin }^{-1}}3/4=\theta $ becomes
${{\sin }^{-1}}3/4=2\theta $ ---- (1)
Now by multiplying sin on both sides of the equation(1)
$\sin ({{\sin }^{-1}}3/4)=\sin 2\theta $
$\sin 2\theta =3/4$ ---- (2)
Now by using the formula $\sin 2\theta =2\tan \theta /(1+{{\tan }^{2}}\theta )$ on equation (2)
$2\tan \theta /(1+{{\tan }^{2}}\theta )=3/4$
Now by cross multiplication
$2\tan \theta /(1+{{\tan }^{2}}\theta )=3/4$ becomes
$4(2\tan \theta )=3(1+{{\tan }^{2}}\theta )$ ----- (3)
By solving the brackets of equation (3)
$8\tan \theta =3+3{{\tan }^{2}}\theta $
$3{{\tan }^{2}}\theta -8\tan \theta +3=0$ ------- (4)
To find the roots of a quadratic equation we use the formula
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
So by using the above formula on equation (4)
$\tan \theta =-(-8)\pm \sqrt{64-(4 \times 3 \times 3}/2 \times 3$
So, $\tan \theta =4\pm \sqrt{7}/3$ ----- (5)
Now by taking tan inverse on both sides of equation (5)
$\theta ={{\tan }^{-1}}\left[ 4\pm \sqrt{7}/3 \right]$
$\tan \theta =4\pm \sqrt{7}/3$
As $\theta =1/2{{\sin }^{-1}}3/4$
So, $\tan \theta =4\pm \sqrt{7}/3$
Since ,
$\begin{align}
& -\pi /2\le {{\sin }^{-1}}3/4\le \pi /2 \\
& -\pi /4\le 1/2{{\sin }^{-1}}3/4\le \pi /4 \\
\end{align}$
Therefore, $\tan (-\pi /4)\le \tan 1/2({{\sin }^{-1}}3/4)\le \tan \pi /4$
$-1\le \tan (1/2{{\sin }^{-1}}3/4)\le 1$
Since, $4+\sqrt{7}/3>1$ so it is ignored
Therefore, $\tan (1/2{{\sin }^{-1}}3/4)=4-\sqrt{7}/3$.
Note:
There is a restriction on $\sin \theta $ i.e. $-\pi /2\le {{\sin }^{-1}}\theta \le \pi /2$. So all the values which are greater than one should be ignored. Always use the correct trigonometric formula to solve a particular equation as using the wrong formula leads towards the wrong answer.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 4 Maths: Engaging Questions & Answers for Success

Trending doubts
Give 10 examples of unisexual and bisexual flowers

Draw a labelled sketch of the human eye class 12 physics CBSE

a Tabulate the differences in the characteristics of class 12 chemistry CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Why is the cell called the structural and functional class 12 biology CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE
