Answer

Verified

450k+ views

Hint: We have to prove that $2{{\tan }^{-1}}\left( \sqrt{\dfrac{a-b}{a+b}}\tan \dfrac{\theta }{2} \right)={{\cos }^{-1}}\left( \dfrac{a\cos \theta +b}{a+b\cos \theta } \right)$ , for that you should prove LHS$=$RHS. Assume $A={{\cos }^{-1}}\left( \dfrac{a\cos \theta +b}{a+b\cos \theta } \right)$ , and use $\cos A=\dfrac{1-{{\tan }^{2}}\dfrac{A}{2}}{1+{{\tan }^{2}}\dfrac{A}{2}}$ and ${{\tan }^{2}}\dfrac{A}{2}=\dfrac{1-\cos A}{1+\cos A}$ .

Try it, you will get the answer.

Complete step-by-step answer:

The trigonometric functions (also called circular functions, angle functions, or trigonometric functions) are real functions that relate an angle of a right-angled triangle to ratios of two side lengths. They are widely used in all sciences that are related to geometry, such as navigation, solid mechanics, celestial mechanics, geodesy, and many others. They are among the simplest periodic functions, and as such are also widely used for studying periodic phenomena, through Fourier analysis.

The most widely used trigonometric functions are the sine, the cosine, and the tangent. Their reciprocals are respectively the cosecant, the secant, and the cotangent, which are less used in modern mathematics.

The cosine function, along with sine and tangent, is one of the three most common trigonometric functions. In any right triangle, the cosine of an angle is the length of the adjacent side (A) divided by the length of the hypotenuse (H). In a formula, it is written simply as '$\cos $'. $\cos $ function (or cosine function) in a triangle is the ratio of the adjacent side to that of the hypotenuse. The cosine function is one of the three main primary trigonometric functions and it is itself the complement of sine (cos + sine).

The cosine graph or the cos graph is an up-down graph just like the sine graph. The only difference between the sine graph and the cos graph is that the sine graph starts from $0$ while the cos graph starts from $90{}^\circ $ (or $\dfrac{\pi }{2}$).

Now we are given $2{{\tan }^{-1}}\left( \sqrt{\dfrac{a-b}{a+b}}\tan \dfrac{\theta }{2} \right)={{\cos }^{-1}}\left( \dfrac{a\cos \theta +b}{a+b\cos \theta } \right)$ .

For that we have to prove LHS$=$RHS.

So first let us consider,

$A={{\cos }^{-1}}\left( \dfrac{a\cos \theta +b}{a+b\cos \theta } \right)$ ……………….. (1)

$\cos A=\left( \dfrac{a\cos \theta +b}{a+b\cos \theta } \right)$

We know that,

$\cos A=\dfrac{1-{{\tan }^{2}}\dfrac{A}{2}}{1+{{\tan }^{2}}\dfrac{A}{2}}$ and ${{\tan }^{2}}\dfrac{A}{2}=\dfrac{1-\cos A}{1+\cos A}$

So now applying these we get,

\[\begin{align}

& {{\tan }^{2}}\dfrac{A}{2}=\dfrac{1-\left( \dfrac{a\cos \theta +b}{a+b\cos \theta } \right)}{1+\left( \dfrac{a\cos \theta +b}{a+b\cos \theta } \right)} \\

& {{\tan }^{2}}\dfrac{A}{2}=\dfrac{a+b\cos \theta -a\cos \theta -b}{a+b\cos \theta +a\cos \theta +b} \\

& {{\tan }^{2}}\dfrac{A}{2}=\dfrac{(a-b)(1-\cos \theta )}{(a+b)(1+\cos \theta )} \\

& {{\tan }^{2}}\dfrac{A}{2}=\dfrac{(a-b)}{(a+b)}{{\tan }^{2}}\dfrac{\theta }{2} \\

\end{align}\]

Now taking square root we get,

\[\begin{align}

& \tan \dfrac{A}{2}=\sqrt{\dfrac{(a-b)}{(a+b)}}{{\tan }^{2}}\dfrac{\theta }{2} \\

& \dfrac{A}{2}={{\tan }^{-1}}\left( \sqrt{\dfrac{(a-b)}{(a+b)}}{{\tan }^{2}}\dfrac{\theta }{2} \right) \\

\end{align}\]

So we get,

\[A=2{{\tan }^{-1}}\left( \sqrt{\dfrac{(a-b)}{(a+b)}}{{\tan }^{2}}\dfrac{\theta }{2} \right)\] ……….. (2)

Now from (1) and (2), we can see that,

\[A={{\cos }^{-1}}\left( \dfrac{a\cos \theta +b}{a+b\cos \theta } \right)=2{{\tan }^{-1}}\left( \sqrt{\dfrac{(a-b)}{(a+b)}}{{\tan }^{2}}\dfrac{\theta }{2} \right)\]

So we get,

$2{{\tan }^{-1}}\left( \sqrt{\dfrac{a-b}{a+b}}\tan \dfrac{\theta }{2} \right)={{\cos }^{-1}}\left( \dfrac{a\cos \theta +b}{a+b\cos \theta } \right)$ .

Hence proved.

Note: Read the question and see what is asked. Your concept regarding trigonometric functions should be clear. A proper assumption should be made. Do not make silly mistakes while substituting. Equate it in a proper manner and don't confuse yourself.

Try it, you will get the answer.

Complete step-by-step answer:

The trigonometric functions (also called circular functions, angle functions, or trigonometric functions) are real functions that relate an angle of a right-angled triangle to ratios of two side lengths. They are widely used in all sciences that are related to geometry, such as navigation, solid mechanics, celestial mechanics, geodesy, and many others. They are among the simplest periodic functions, and as such are also widely used for studying periodic phenomena, through Fourier analysis.

The most widely used trigonometric functions are the sine, the cosine, and the tangent. Their reciprocals are respectively the cosecant, the secant, and the cotangent, which are less used in modern mathematics.

The cosine function, along with sine and tangent, is one of the three most common trigonometric functions. In any right triangle, the cosine of an angle is the length of the adjacent side (A) divided by the length of the hypotenuse (H). In a formula, it is written simply as '$\cos $'. $\cos $ function (or cosine function) in a triangle is the ratio of the adjacent side to that of the hypotenuse. The cosine function is one of the three main primary trigonometric functions and it is itself the complement of sine (cos + sine).

The cosine graph or the cos graph is an up-down graph just like the sine graph. The only difference between the sine graph and the cos graph is that the sine graph starts from $0$ while the cos graph starts from $90{}^\circ $ (or $\dfrac{\pi }{2}$).

Now we are given $2{{\tan }^{-1}}\left( \sqrt{\dfrac{a-b}{a+b}}\tan \dfrac{\theta }{2} \right)={{\cos }^{-1}}\left( \dfrac{a\cos \theta +b}{a+b\cos \theta } \right)$ .

For that we have to prove LHS$=$RHS.

So first let us consider,

$A={{\cos }^{-1}}\left( \dfrac{a\cos \theta +b}{a+b\cos \theta } \right)$ ……………….. (1)

$\cos A=\left( \dfrac{a\cos \theta +b}{a+b\cos \theta } \right)$

We know that,

$\cos A=\dfrac{1-{{\tan }^{2}}\dfrac{A}{2}}{1+{{\tan }^{2}}\dfrac{A}{2}}$ and ${{\tan }^{2}}\dfrac{A}{2}=\dfrac{1-\cos A}{1+\cos A}$

So now applying these we get,

\[\begin{align}

& {{\tan }^{2}}\dfrac{A}{2}=\dfrac{1-\left( \dfrac{a\cos \theta +b}{a+b\cos \theta } \right)}{1+\left( \dfrac{a\cos \theta +b}{a+b\cos \theta } \right)} \\

& {{\tan }^{2}}\dfrac{A}{2}=\dfrac{a+b\cos \theta -a\cos \theta -b}{a+b\cos \theta +a\cos \theta +b} \\

& {{\tan }^{2}}\dfrac{A}{2}=\dfrac{(a-b)(1-\cos \theta )}{(a+b)(1+\cos \theta )} \\

& {{\tan }^{2}}\dfrac{A}{2}=\dfrac{(a-b)}{(a+b)}{{\tan }^{2}}\dfrac{\theta }{2} \\

\end{align}\]

Now taking square root we get,

\[\begin{align}

& \tan \dfrac{A}{2}=\sqrt{\dfrac{(a-b)}{(a+b)}}{{\tan }^{2}}\dfrac{\theta }{2} \\

& \dfrac{A}{2}={{\tan }^{-1}}\left( \sqrt{\dfrac{(a-b)}{(a+b)}}{{\tan }^{2}}\dfrac{\theta }{2} \right) \\

\end{align}\]

So we get,

\[A=2{{\tan }^{-1}}\left( \sqrt{\dfrac{(a-b)}{(a+b)}}{{\tan }^{2}}\dfrac{\theta }{2} \right)\] ……….. (2)

Now from (1) and (2), we can see that,

\[A={{\cos }^{-1}}\left( \dfrac{a\cos \theta +b}{a+b\cos \theta } \right)=2{{\tan }^{-1}}\left( \sqrt{\dfrac{(a-b)}{(a+b)}}{{\tan }^{2}}\dfrac{\theta }{2} \right)\]

So we get,

$2{{\tan }^{-1}}\left( \sqrt{\dfrac{a-b}{a+b}}\tan \dfrac{\theta }{2} \right)={{\cos }^{-1}}\left( \dfrac{a\cos \theta +b}{a+b\cos \theta } \right)$ .

Hence proved.

Note: Read the question and see what is asked. Your concept regarding trigonometric functions should be clear. A proper assumption should be made. Do not make silly mistakes while substituting. Equate it in a proper manner and don't confuse yourself.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE