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Prove that \[\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{1}/\mathbf{4}\text{ }+\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{2}/\mathbf{9}\text{ }=\text{ }\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{1}/\surd \mathbf{5}\]
Prove that \[\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{1}/\mathbf{4}\text{ }+\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{2}/\mathbf{9}\text{ }=\text{ }\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{1}/\surd \mathbf{5}\]
CBSE | Class 12 | Exercise 1 | Inverse Trigonometric Functions | NCERT Exemplar
Prove that \[\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{1}/\mathbf{4}\text{ }+\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{2}/\mathbf{9}\text{ }=\text{ }\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{1}/\surd \mathbf{5}\]

Taking the LHS,

\[ta{{n}^{-1}}~1/4\text{ }+\text{ }ta{{n}^{-1}}~2/9\]

= RHS

– Hence Proved

i

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