Question

Asec alpha - ctan alpha = d and bsec alpha + dtan alpha=d then

Answer 1

Hope this will help u..................

Answer 2

asec alpha - ctan alpha = d and bsec alpha - dtan alpha=c

$a \sec \alpha - c \tan \alpha = d \\ a \sec \alpha = d + c \tan \alpha \\ squaring \: on \: both \: sides \\ {a}^{2} \sec^{2} \alpha = {d}^{2} + {c}^{2}tan^{2} \alpha \: \: \: \: \: \: \: \: .....(1) \\ now \\ b \: sec \alpha \: - d \: tan \alpha = c \\ b \: sec \alpha = c + d \: tan \alpha \\ squaring \: on \: both \: sides \: \\{b}^{2} sec ^{2} \alpha = {c}^{2} + {d}^{2} tan^{2} \alpha \: \: \: \: ...... (2) \\ adding \: (1) \: and \: (2) \\( {a}^{2} + {b}^{2} )sec ^{2} \alpha = ({c}^{2} + {d}^{2}) + ({c}^{2} + {d}^{2}) {tan}^{2} \alpha \\( {a}^{2} + {b}^{2} )sec ^{2} \alpha = ({c}^{2} + {d}^{2})(1 + tan ^{2} \alpha ) \\( {a}^{2} + {b}^{2} )sec ^{2} \alpha = ({c}^{2} + {d}^{2})(sec ^{2} \alpha ) \\ {a}^{2} + {b}^{2} = {c}^{2} + {d}^{2}$