{"id":11615,"date":"2022-08-02T14:24:34","date_gmt":"2022-08-02T08:54:34","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11615"},"modified":"2022-08-02T14:24:42","modified_gmt":"2022-08-02T08:54:42","slug":"in-the-figure-find-tan-p-cot-r","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/in-the-figure-find-tan-p-cot-r\/","title":{"rendered":"In the figure, find tan P – cot R ?"},"content":{"rendered":"
In triangle PQR, we have<\/p>\n
\\(\\angle\\) Q = 90,\u00a0 PQ = 12 cm and PR = 13 cm<\/p>\n
Using Pythagoras Theorem,<\/p>\n
\\({PR}^2\\) = \\({PQ}^2\\) + \\({QR}^2\\)<\/p>\n
\\(\\implies\\)\u00a0 \\({QR}^2\\) = \\({PR}^2\\) – \\({PQ}^2\\)<\/p>\n
= \\((13)^2\\) – \\((12)^2\\) = 25<\/p>\n
\\(\\implies\\)\u00a0 QR = 5 cm<\/p>\n
We know that,<\/p>\n
tan P = \\(QR\\over PQ\\) = tan P = \\(5\\over 12\\)\u00a0 \u00a0 \u00a0 \u00a0……..(1)<\/p>\n
cot R = \\(QR\\over PQ\\)\u00a0 \\(\\implies\\)\u00a0 cot R = \\(5\\over 12\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ………(2)<\/p>\n
From (1) and (2), we have<\/p>\n
tan P – cot R = \\(5\\over 12\\) – \\(5\\over 12\\) = 0<\/p>\n","protected":false},"excerpt":{"rendered":"
Solution : In triangle PQR, we have \\(\\angle\\) Q = 90,\u00a0 PQ = 12 cm and PR = 13 cm Using Pythagoras Theorem, \\({PR}^2\\) = \\({PQ}^2\\) + \\({QR}^2\\) \\(\\implies\\)\u00a0 \\({QR}^2\\) = \\({PR}^2\\) – \\({PQ}^2\\) = \\((13)^2\\) – \\((12)^2\\) = 25 \\(\\implies\\)\u00a0 QR = 5 cm We know that, tan P = \\(QR\\over PQ\\) = tan …<\/p>\n