What is the best solution of ncert class 10 of introduction of trigonometry , how i solve the exercise 8.1 question number 2 easily and simple way . This is very important question of trigonometry . In Fig. 8.13, find tan P – cot R

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In the given triangle PQR, the given triangle is right angled at Q and the given measures are:

PR = 13cm,

PQ = 12cm

Since the given triangle is right angled triangle, to find the side QR, apply the Pythagorean theorem

According to Pythagorean theorem,

In a right- angled triangle, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides.

PR

^{2}= QR^{2}+ PQ^{2}Substitute the values of PR and PQ

13

^{2 }= QR^{2}+12^{2}169 = QR

^{2}+144Therefore, QR

^{2 }= 169−144QR

^{2 }= 25QR = √25 = 5

Therefore, the side QR = 5 cm

To find tan P – cot R:

According to the trigonometric ratio, the tangent function is equal to the ratio of the length of the opposite side to the adjacent sides, the value of tan (P) becomes

tan (P)

=Opposite side /Adjacent side = QR/PQ = 5/12Since cot function is the reciprocal of the tan function, the ratio of cot function becomes,

Cot (R) = Adjacent side/Opposite side = QR/PQ = 5/12

Therefore,

tan (P) – cot (R) = 5/12 – 5/12 = 0

Therefore, tan(P) – cot(R) = 0