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Prove that \[\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{8}/\mathbf{17}\text{ }+\text{ }\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{3}/\mathbf{5}\text{ }=\text{ }\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{77}/\mathbf{85}\]
Prove that \[\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{8}/\mathbf{17}\text{ }+\text{ }\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{3}/\mathbf{5}\text{ }=\text{ }\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{77}/\mathbf{85}\]
CBSE | Class 12 | Exercise 1 | Inverse Trigonometric Functions | Maths | NCERT Exemplar
Prove that \[\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{8}/\mathbf{17}\text{ }+\text{ }\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{3}/\mathbf{5}\text{ }=\text{ }\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{77}/\mathbf{85}\]

Taking the L.H.S,

= \[\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{8}/\mathbf{17}\text{ }+\text{ }\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{3}/\mathbf{5}\]

= tan-1 8/15 + tan-1 3/4

– Hence proved

i

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