Math, asked by shushmitha21, 3 months ago

prove that (1+tan A - secA) (1+cotA+cosecA)=2

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Answered by sandy1816

$(1 + tana - seca)(1 + cota + coseca) \\ \\ = ( \frac{cosa + sina - 1}{cosa} )( \frac{sina + cosa + 1}{sina} ) \\ \\ = \frac{( {cosa + sina})^{2} - 1 }{cosasina} \\ \\ = \frac{1 + 2sinacosa - 1}{sinacosa} \\ \\ = 2$

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prove that (1+tan A - secA) (1+cotA+cosecA)=2​

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prove that (1+tan A - secA) (1+cotA+cosecA)=2