If A=35^ ^ ^ , B=15^ ^ ^ and C=40^ ^ ^ , then A · B+ B · C+ C · A is equal to

Correct answer: B. 1 Given that, A=35^ ^ ^ , B=15^ ^ ^ and C=40^ ^ ^ aligned & (A+B+C) & = [ array c A+ B+ C - A B C array ] [ array c 1- A B- B C - C A array ] & (90^ ^ ^ ) & =( A+ B+ C- A B C)/(1- A B- B C- C A) & A B+ B C & + C A=1 & aligned

JEE Main

Exam: JEE Main
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

JEE Main · Mathematics · Trigonometric Ratios & Identities

If A=35^ ^ ^ , B=15^ ^ ^ and C=40^ ^ ^ , then A · B+ B · C+ C · A...

A=35^{{}^{{}^{\circ}}}, B=15^{{}^{{}^{\circ}}}

and

C=40^{{}^{{}^{\circ}}}

, then

\tan A \cdot \tan B+\tan B \cdot \tan C+\tan C \cdot \tan A

is equal to

Options

A $0$
B $1$
C $2$
D $3$

Correct answer

B. $1$

Step-by-step solution

Given that,

A=35^{{}^{{}^{\circ}}}, B=15^{{}^{{}^{\circ}}}

and

C=40^{{}^{{}^{\circ}}}

\begin{aligned}& \therefore \tan (A+B+C) \\ & =\frac{\left[\begin{array}{c} \tan A+\tan B+\tan C \\ -\tan A \tan B \tan C \end{array}\right]}{\left[\begin{array}{c} 1-\tan A \tan B-\tan B \tan C \\ -\tan C \tan A \end{array}\right]}\\ & \Rightarrow \tan \left(90^{{}^{{}^{\circ}}}\right) \\ & =\frac{\tan A+\tan B+\tan C-\tan A \tan B \tan C}{1-\tan A \tan B-\tan B \tan C-\tan C \tan A}\\ & \Rightarrow \quad \tan A \tan B+\tan B \tan C \\ & +\tan C \tan A=1 \\ & \end{aligned}

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