10th Class Mathematics Solved Paper - Mathematics-2015 Term-I

  • question_answer Prove that: \[(1+cot\text{ }A+tan\text{ }A).\text{(}sin\text{ }A-cos\text{ }A)\]\[=\frac{{{\sec }^{3}}A-\cos e{{c}^{3}}A}{{{\sec }^{2}}A.\cos e{{c}^{2}}A}\]

    Answer:

    \[L.H.S.=(1+cot\text{ }A+tan\text{ }A)\text{ (}\sin \text{ }A-cos\text{ }A)\]
    \[=\left( 1+\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A} \right)(sin\,A-cos\,A)\]
    \[=\left( \frac{\sin \,A\cos A+{{\cos }^{2}}A+{{\sin }^{2}}A}{\sin A.\cos A} \right)(sinA-cosA)\]
    \[=\frac{{{\sin }^{3}}A-{{\cos }^{3}}A}{\sin A.\cos A}\]   
                            \[[\text{Using}\,{{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+ab+{{b}^{2}})\]
    \[=\frac{\frac{{{\sin }^{3}}A}{{{\sin }^{3}}A.{{\cos }^{3}}A}-\frac{{{\cos }^{3}}A}{{{\sin }^{3}}A.{{\cos }^{3}}A}}{\frac{\sin A\cos A}{{{\sin }^{3}}A.{{\cos }^{3}}A}}\]
                                     [Dividing Num. & Denom. by \[si{{n}^{3}}A.\,co{{s}^{3}}A\] ]
    \[=\frac{{{\sec }^{3}}A-\cos e{{c}^{3}}A}{{{\sec }^{2}}A.\cos e{{c}^{2}}A}=R.H.S.\]                   Hence Proved.


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