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Home » If cos A = –12/13 and cot B = 24/7, where A lies in the second quadrant and B in the third quadrant, find the values of the following: tan (A + B)
If cos A = –12/13 and cot B = 24/7, where A lies in the second quadrant and B in the third quadrant, find the values of the following: tan (A + B)
CBSE | Class 11 | Exercise 7.1 | Maths | RD Sharma | Values of Trigonometric Functions at Sum or Difference of Angles
If cos A = –12/13 and cot B = 24/7, where A lies in the second quadrant and B in the third quadrant, find the values of the following: tan (A + B)

    \[cos\text{ }A\text{ }=\text{ }-12/13\text{ }and\text{ }cot\text{ }B\text{ }=\text{ }24/7\]

Since, A lies in second quadrant, B in the third quadrant.

Sine function is positive, in the second quadrant.

Both sine and cosine functions are negative, in the third quadrant.

By using the formulas,

    \[sin\text{ }A\text{ }=\text{ }\surd (1\text{ }-\text{ }co{{s}^{2}}~A),\text{ }sin\text{ }B\text{ }=\text{ }\text{ }1/\surd (1\text{ }+\text{ }co{{t}^{2}}~B)\]

And

    \[cos\text{ }B\text{ }=\text{ }-\surd (1\text{ }-\text{ }si{{n}^{2}}~B),\]

Let’s get the value of

    \[sin\text{ }A\text{ }and\text{ }sin\text{ }B\]

    \[sin\text{ }A\text{ }=\text{ }\surd (1\text{ }-\text{ }co{{s}^{2}}~A)\]

    \[=\text{ }\surd (1\text{ }-\text{ }{{\left( -12/13 \right)}^{2}})\]

    \[=\text{ }\surd \left( 1\text{ }-\text{ }144/169 \right)\]

    \[=\text{ }\surd \left( \left( 169-144 \right)/169 \right)\]

    \[=\text{ }\surd \left( 25/169 \right)\]

    \[=\text{ }5/13\]

Or,

    \[sin\text{ }B\text{ }=\text{ }\text{ }1/\surd (1\text{ }+\text{ }co{{t}^{2}}~B)\]

    \[=-\text{ }\text{ }1/\surd (1\text{ }+\text{ }{{\left( 24/7 \right)}^{2}})\]

    \[=-\text{ }\text{ }1/\surd \left( 1\text{ }+\text{ }576/49 \right)\]

    \[=\text{ }-1/\surd \left( \left( 49+576 \right)/49 \right)\]

    \[=\text{ }-1/\surd \left( 625/49 \right)\]

    \[=\text{ }-1/\left( 25/7 \right)\]

    \[=\text{ }-7/25\]

Or,

    \[cos\text{ }B\text{ }=\text{ }-\surd (1\text{ }-\text{ }si{{n}^{2}}~B)\]

    \[=\text{ }-\surd (1-{{\left( -7/25 \right)}^{2}})\]

    \[=\text{ }-\surd \left( 1-\left( 49/625 \right) \right)\]

    \[=\text{ }-\surd \left( \left( 625-49 \right)/625 \right)\]

    \[=\text{ }-\surd \left( 576/625 \right)\]

    \[=\text{ }-24/25\]

Therefore, now let us find

 

    \[tan\text{ }\left( A\text{ }+\text{ }B \right)\]

Since,

    \[tan\text{ }\left( A\text{ }+\text{ }B \right)\text{ }=\text{ }sin\text{ }\left( A+B \right)\text{ }/\text{ }cos\text{ }\left( A+B \right)\]

    \[=\text{ }\left( -36/325 \right)\text{ }/\text{ }\left( 323/325 \right)\]

    \[=\text{ }-36/323\]

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