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

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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.

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