In the figure, find tan P – cot R ?

Solution :

In triangle PQR, we havetriangle

\(\angle\) Q = 90,  PQ = 12 cm and PR = 13 cm

Using Pythagoras Theorem,

\({PR}^2\) = \({PQ}^2\) + \({QR}^2\)

\(\implies\)  \({QR}^2\) = \({PR}^2\) – \({PQ}^2\)

= \((13)^2\) – \((12)^2\) = 25

\(\implies\)  QR = 5 cm

We know that,

tan P = \(QR\over PQ\) = tan P = \(5\over 12\)       ……..(1)

cot R = \(QR\over PQ\)  \(\implies\)  cot R = \(5\over 12\)            ………(2)

From (1) and (2), we have

tan P – cot R = \(5\over 12\) – \(5\over 12\) = 0

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