In the given figure, ABC is an isosceles triangle with BC = 8 cm and AB = AC = 5 cm. The value of tan C - cot B is ________.

Given:

ΔABC is an isosceles triangle

AB = AC = 5 cm and BC = 8 cm

Concept Used:

In an isosceles triangle, the median to the base from the opposite vertex and the perpendicular bisector of the base are the same.

Calculation:

In ΔABC, AD is perpendicular bisector

⇒ BD = DC = BC/2 = 8/2 = 4 cm

In ΔADC, by Pythagoras Theorem,

AC² = DC² + AD²

⇒ AD² = 5² - 4² = 9

⇒ AD = 3 cm

Now, in ΔADC,

tan C = AD/DC = 3/4

Now, in ΔADB,

cot B = BD/AD = 4/3

⇒ tan C - cot B = (3/4) - (4/3)

⇒ tan C - cot B = (9 - 16)/12

⇒ tan C - cot B = (-7)/12

∴ The value of tan C - cot B is (-7)/12.

Alternate Method

Given:

In the given figure, ABC is an isosceles triangle with BC = 8 cm and AB = AC = 5 cm.

Concept used:

1. sin² α + cos2 α = 1

2. tan α = sin α ÷ cos α = 1/ cot α

3. In trigonometry, the Cosine Rule says that the square of the length of any side of a given triangle is equal to the sum of the squares of the length of the other sides minus twice the product of the other two sides multiplied by the cosine of angle included between them.

Calculation:

​According to the concept (Cosine Rule),

cos B = \(\frac {8^2 + 5^2 - 5^2}{2 \times 8 \times 5}\) = \(\frac {4}{5}\)      ....(1)

So, sin B

⇒ \(\sqrt {1 - cos^2 B}\)

⇒ \(\sqrt {1 - ({\frac {4}{5}})^2}\)

⇒ \(\sqrt {1 - {\frac {16}{25}}}\)

⇒ \(\sqrt {{\frac {25 - 16}{25}}}\)

⇒ \(\sqrt {\frac {9}{25}}\)

⇒ \(\frac {3}{5}\)      ....(2)

Now, tan C - cot B

⇒ tan B - cot B (∵ ABC is an isosceles triangle where AB = AC. So, ∠B = ∠C)

⇒ \(\frac {sin B}{cos B} - \frac {cos B}{sin B}\)

⇒ \(\frac {sin^2 B - cos^2 B}{sin B \times cos B} \)

⇒ \(\frac {(\frac {3}{5})^2 - (\frac {4}{5})^2}{\frac {3}{5} \times \frac {4}{5}}\) (From 1 and 2)

⇒ \(-\frac{7}{12}\)

∴ The value of tan C - cot B is \(-\frac{7}{12}\).