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3. In fig. 5.37, find \[\mathbf{tan}\text{ }\mathbf{P}\]and\[\mathbf{cot}\text{ }\mathbf{R}\]. Is \[\mathbf{tan}\text{ }\mathbf{P}\text{ }=\text{ }\mathbf{cot}\text{ }\mathbf{R}?~\]
3. In fig. 5.37, find \[\mathbf{tan}\text{ }\mathbf{P}\]and\[\mathbf{cot}\text{ }\mathbf{R}\]. Is \[\mathbf{tan}\text{ }\mathbf{P}\text{ }=\text{ }\mathbf{cot}\text{ }\mathbf{R}?~\]
CBSE | Class 10 | Maths | RD Sharma | Trigonometric Ratios
3. In fig. 5.37, find \[\mathbf{tan}\text{ }\mathbf{P}\]and\[\mathbf{cot}\text{ }\mathbf{R}\]. Is \[\mathbf{tan}\text{ }\mathbf{P}\text{ }=\text{ }\mathbf{cot}\text{ }\mathbf{R}?~\]

Solution:

By using Pythagoras theorem in \[\vartriangle PQR\], we have

\[P{{R}^{2}}~=\text{ }P{{Q}^{2}}~+\text{ }Q{{R}^{2}}\]

Putting the length of given side PR and PQ in the above equation

\[{{13}^{2~}}=\text{ }{{12}^{2}}~+\text{ }Q{{R}^{2}}\]

\[Q{{R}^{2}}~=\text{ }{{13}^{2}}~\text{ }{{12}^{2}}\]

\[Q{{R}^{2}}~=\text{ }169\text{ }\text{ }144\]

\[Q{{R}^{2~}}=\text{ }25\]

\[QR\text{ }=~\surd 25\text{ }=\text{ }5\]

By definition,

tan P = Perpendicular side opposite to P/ Base side adjacent to angle P

\[tan\text{ }P\text{ }=\text{ }QR/PQ\]

\[tan\text{ }P\text{ }=\text{ }5/12~\ldots \ldots \ldots .\text{ }\left( 1 \right)\]

And,

cot R= Base/Perpendicular

\[cot\text{ }R=\text{ }QR/PQ\]

\[cot\text{ }R=\text{ }5/12~\ldots .\text{ }\left( 2 \right)\]

When comparing \[equation\text{ }\left( 1 \right)\text{ }and\text{ }\left( 2 \right)\], we can see that R.H.S of both the equation is equal.

Therefore, L.H.S of both equations should also be equal.

\[\therefore tan\text{ }P\text{ }=\text{ }cot\text{ }R\]

Yes, \[\mathbf{tan}\text{ }\mathbf{P}\text{ }=\text{ }\mathbf{cot}\text{ }\mathbf{R}\text{ }=~\mathbf{5}/\mathbf{12}\]

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