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Question: If \[A + B + C = {90^ \circ }\], then find the value of \[\tan A\tan B + \tan B\tan C + \tan C\tan A...

If A+B+C=90∘A + B + C = {90^ \circ }, then find the value of tan⁡Atan⁡B+tan⁡Btan⁡C+tan⁡Ctan⁡A=\tan A\tan B + \tan B\tan C + \tan C\tan A = .

Explanation

Solution

Here we write the sum of angles A and B by shifting the angle C to the other side and then apply function tan on both sides. Use the formula for tan⁡(x+y)\tan (x + y) and the formula tan⁡(90∘−θ)=cot⁡θ\tan ({90^ \circ } - \theta ) = \cot \theta solve to find the value of tan⁡Atan⁡B+tan⁡Btan⁡C+tan⁡Ctan⁡A\tan A\tan B + \tan B\tan C + \tan C\tan A.

  • Tan and cot are complementary to each other, which means if angle inside the tan function is xx then angle inside the cot function will be 90−x90 - x then only the values of tan and cot will become equal, i.e. tan⁡(90∘−θ)=cot⁡θ\tan ({90^ \circ } - \theta ) = \cot \theta and cot⁡(90∘−θ)=tan⁡θ\cot ({90^ \circ } - \theta ) = \tan \theta
  • tan⁡(x+y)=tan⁡x+tan⁡y1−tan⁡xtan⁡y\tan (x + y) = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}

Complete step-by-step answer:
We are given A+B+C=90∘A + B + C = {90^ \circ }
Shift the value of C to right hand side of the equation
⇒A+B=90∘−C\Rightarrow A + B = {90^ \circ } - C
Now we apply the function tan on both sides
⇒tan⁡(A+B)=tan⁡(90∘−C)\Rightarrow \tan (A + B) = \tan ({90^ \circ } - C)
Use the formula tan⁡(90∘−θ)=cot⁡θ\tan ({90^ \circ } - \theta ) = \cot \theta because tan and cot are complementary to each other.
⇒tan⁡(A+B)=cot⁡C\Rightarrow \tan (A + B) = \cot C
Since we know cot⁡θ=1tan⁡θ\cot \theta = \dfrac{1}{{\tan \theta }}
⇒tan⁡(A+B)=1tan⁡C\Rightarrow \tan (A + B) = \dfrac{1}{{\tan C}}
Now we know tan⁡(x+y)=tan⁡x+tan⁡y1−tan⁡xtan⁡y\tan (x + y) = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}, Substitute the value of x as A and y as B.
⇒tan⁡A+tan⁡B1−tan⁡Atan⁡B=1tan⁡C\Rightarrow \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}} = \dfrac{1}{{\tan C}}
Now cross multiply the values from both sides of the equation.
⇒tan⁡C(tan⁡A+tan⁡B)=1−tan⁡Atan⁡B\Rightarrow \tan C(\tan A + \tan B) = 1 - \tan A\tan B
Multiply the terms on the left hand side of the equation.
⇒tan⁡Ctan⁡A+tan⁡Ctan⁡B=1−tan⁡Atan⁡B\Rightarrow \tan C\tan A + \tan C\tan B = 1 - \tan A\tan B
Shift the value of tan⁡Atan⁡B\tan A\tan Bfrom right side of the equation to left side of the equation.
⇒tan⁡Ctan⁡A+tan⁡Ctan⁡B+tan⁡Atan⁡B=1\Rightarrow \tan C\tan A + \tan C\tan B + \tan A\tan B = 1
Writing the terms in proper manner
⇒tan⁡Atan⁡B+tan⁡Btan⁡C+tan⁡Ctan⁡A=1\Rightarrow \tan A\tan B + \tan B\tan C + \tan C\tan A = 1

So, the value of tan⁡Atan⁡B+tan⁡Btan⁡C+tan⁡Ctan⁡A\tan A\tan B + \tan B\tan C + \tan C\tan A is 1.

Note: Students many times make mistakes while shifting the values from one side of the equation to another side of the equation. Keep in mind that sign changes from positive to negative and vice versa when we shift a value from one side to another side. Also, keep in mind that in these kinds of questions we always write the angle as capital letter and we don’t use small letters to denote the angle along with trigonometric function.
Many students try to solve the question by substituting the value of tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} which will make the solution very complex as there are already three angles A, B and C, so try to avoid more complex solution like this.