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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: (i) sin (A + B) (ii) cos (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: (i) sin (A + B) (ii) cos (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: (i) sin (A + B) (ii) cos (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.

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 A and sin 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,

    \[\left( \mathbf{i} \right)~sin\text{ }\left( A\text{ }+\text{ }B \right)\]

Since,

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

Therefore,

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

    \[=\text{ }5/13\text{ }\times \text{ }\left( -24/25 \right)\text{ }+\text{ }\left( -12/13 \right)\text{ }\times \text{ }\left( -7/25 \right)\]

    \[=\text{ }-120/325\text{ }+\text{ }84/325\]

    \[=\text{ }-36/325\]

 

    \[\left( \mathbf{ii} \right)~cos\text{ }\left( A\text{ }+\text{ }B \right)\]

Since,

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

Therefore,

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

    \[=\text{ }-12/13\text{ }\times \text{ }\left( -24/25 \right)\text{ }-\text{ }\left( 5/13 \right)\text{ }\times \text{ }\left( -7/25 \right)\]

    \[=\text{ }288/325\text{ }+\text{ }35/325\]

    \[=\text{ }323/325\]

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