If A, B and C are the angles of a triangle such that A: B: C=2: 3: 4, then what is the value of the following?
\n$$ \\frac{\\tan^2 B+1}{\\tan^2 B-1} $$<\/p>\n
[amp_mcq option1=”4″ option2=”2″ option3=”1″ option4=”0″ correct=”option2″]<\/p>\n
The angles are:
\nA = $2x = 2 \\times 20\u00b0 = 40\u00b0$
\nB = $3x = 3 \\times 20\u00b0 = 60\u00b0$
\nC = $4x = 4 \\times 20\u00b0 = 80\u00b0$<\/p>\n
We need to find the value of the expression $\\frac{\\tan^2 B+1}{\\tan^2 B-1}$.
\nSubstitute $B = 60\u00b0$:
\n$\\tan B = \\tan 60\u00b0 = \\sqrt{3}$.
\n$\\tan^2 B = (\\sqrt{3})^2 = 3$.<\/p>\n
Now substitute this value into the expression: If A, B and C are the angles of a triangle such that A: B: C=2: 3: 4, then what is the value of the following? $$ \\frac{\\tan^2 B+1}{\\tan^2 B-1} $$ [amp_mcq option1=”4″ option2=”2″ option3=”1″ option4=”0″ correct=”option2″] This question was previously asked in UPSC CBI DSP LDCE – 2023 Download PDFAttempt Online Let the angles … <\/p>\n
\n$\\frac{\\tan^2 B+1}{\\tan^2 B-1} = \\frac{3+1}{3-1} = \\frac{4}{2} = 2$.
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\n$\\frac{\\tan^2 B+1}{\\tan^2 B-1} = \\frac{\\sec^2 B}{\\frac{\\sin^2 B}{\\cos^2 B}-1} = \\frac{\\frac{1}{\\cos^2 B}}{\\frac{\\sin^2 B – \\cos^2 B}{\\cos^2 B}} = \\frac{1}{\\sin^2 B – \\cos^2 B}$.
\nUsing the identity $\\cos(2B) = \\cos^2 B – \\sin^2 B = -(\\sin^2 B – \\cos^2 B)$, the expression is $\\frac{1}{-\\cos(2B)}$.
\nFor $B=60\u00b0$, $2B=120\u00b0$. $\\cos(120\u00b0) = -\\frac{1}{2}$.
\nThe value is $\\frac{1}{-(-\\frac{1}{2})} = \\frac{1}{\\frac{1}{2}} = 2$.
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