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Question: The value of \( \sec \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{b + a}}{{b - a}}} \right) - {{\tan }^{ ...

The value of sec⁡[tan⁡−1(b+ab−a)−tan⁡−1(ab)]\sec \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{b + a}}{{b - a}}} \right) - {{\tan }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right] is ______________.

Explanation

Solution

Hint : Here we are given two different trigonometric functions, secant function and tangent inverse function; there are different formulas that can help us find the relation between them. We see that the secant function lies outside all the brackets and inside the brackets we have two inverse tangent functions. Always simplify the inner bracket first.
Formulas used:
tan⁡−1(x)−tan⁡−1(y)=tan⁡−1(x−y1+xy){\tan ^{ - 1}}(x) - {\tan ^{ - 1}}(y) = {\tan ^{ - 1}}\left( {\dfrac{{x - y}}{{1 + xy}}} \right)
tan⁡−1(1)=π4{\tan ^{ - 1}}(1) = \dfrac{\pi }{4}

Complete step-by-step answer :
The question given to us requires us to find the value of this trigonometric form: sec⁡[tan⁡−1(b+ab−a)−tan⁡−1(ab)]\sec \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{b + a}}{{b - a}}} \right) - {{\tan }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right]
We can solve this question in two parts;
First part we can solve the terms within the inner bracket that is:
[tan⁡−1(b+ab−a)−tan⁡−1(ab)]\left[ {{{\tan }^{ - 1}}\left( {\dfrac{{b + a}}{{b - a}}} \right) - {{\tan }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right]
Once we simplify the inner terms, we can move forward with finding: sec⁡[tan⁡−1(b+ab−a)−tan⁡−1(ab)]\sec \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{b + a}}{{b - a}}} \right) - {{\tan }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right]
To solve the first part we can use a formula that involves the subtraction of inverse of tangents, the formula looks like this:
⇒tan⁡−1(x)−tan⁡−1(y)=tan⁡−1(x−y1+xy)\Rightarrow {\tan ^{ - 1}}(x) - {\tan ^{ - 1}}(y) = {\tan ^{ - 1}}\left( {\dfrac{{x - y}}{{1 + xy}}} \right) ; here the values of xx and yy are given within the question.
⇒x=b+ab−a\Rightarrow x = \dfrac{{b + a}}{{b - a}} and ⇒y=ab\Rightarrow y = \dfrac{a}{b}
Next we can proceed with applying the formula, so we substitute the values in the question in place of xx and yy :
⇒tan⁡−1(b+ab−a)−tan⁡−1(ab)=tan⁡−1((b+ab−a)−(ab)1+(b+ab−a×ab))\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{b + a}}{{b - a}}} \right) - {\tan ^{ - 1}}\left( {\dfrac{a}{b}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{\left( {\dfrac{{b + a}}{{b - a}}} \right) - \left( {\dfrac{a}{b}} \right)}}{{1 + \left( {\dfrac{{b + a}}{{b - a}} \times \dfrac{a}{b}} \right)}}} \right)
Then we simplify the terms within the brackets;
⇒tan⁡−1(b+ab−a)−tan⁡−1(ab)=tan⁡−1(b2+ab−ab+a2b2−ab+ab+a2)\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{b + a}}{{b - a}}} \right) - {\tan ^{ - 1}}\left( {\dfrac{a}{b}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{{b^2} + ab - ab + {a^2}}}{{{b^2} - ab + ab + {a^2}}}} \right)
Simplifying further we get;
⇒tan⁡−1(b+ab−a)−tan⁡−1(ab)=tan⁡−1(b2+a2b2+a2)\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{b + a}}{{b - a}}} \right) - {\tan ^{ - 1}}\left( {\dfrac{a}{b}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{{b^2} + {a^2}}}{{{b^2} + {a^2}}}} \right)
⇒tan⁡−1(b+ab−a)−tan⁡−1(ab)=tan⁡−1(1)\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{b + a}}{{b - a}}} \right) - {\tan ^{ - 1}}\left( {\dfrac{a}{b}} \right) = {\tan ^{ - 1}}\left( 1 \right)
This can also be written as;
⇒tan⁡−1(b+ab−a)−tan⁡−1(ab)=π4\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{b + a}}{{b - a}}} \right) - {\tan ^{ - 1}}\left( {\dfrac{a}{b}} \right) = \dfrac{\pi }{4}
Now that we have simplified the inner bracket to its maximum, we can write the term that was there outside the bracket;
⇒sec⁡[tan⁡−1(b+ab−a)−tan⁡−1(ab)]=sec⁡(π4)\Rightarrow \sec \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{b + a}}{{b - a}}} \right) - {{\tan }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right] = \sec \left( {\dfrac{\pi }{4}} \right)
⇒sec⁡[tan⁡−1(b+ab−a)−tan⁡−1(ab)]=1cos⁡(π4)\Rightarrow \sec \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{b + a}}{{b - a}}} \right) - {{\tan }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right] = \dfrac{1}{{\cos \left( {\dfrac{\pi }{4}} \right)}}
Then we obtain the answer:
⇒sec⁡[tan⁡−1(b+ab−a)−tan⁡−1(ab)]=2\Rightarrow \sec \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{b + a}}{{b - a}}} \right) - {{\tan }^{ - 1}}\left( {\dfrac{a}{b}} \right)} \right] = \sqrt 2
Therefore the answer is the final value we obtain for the given trigonometric form, it is 2\sqrt 2 .
So, the correct answer is “ 2\sqrt 2 ”.

Note : We should know the values of basic trigonometric angular function. We should be thorough with at least values of 0∘,30∘,45∘,60∘,90∘{0^ \circ },{30^ \circ },{45^ \circ },{60^ \circ },{90^ \circ } for the trigonometric functions sine, cosine, tangent. The values for other functions like secant, cosecant and cotangent can be calculated by taking the inverted values of cosine, sine and tangent functions respectively.