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Question
Without using trigonometric identity , show that :
`sin42^circ sec48^circ + cos42^circ cosec48^circ = 2`
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Solution
`sin42^circ sec48^circ + cos42^circ cosec48^circ = 2`
consider `sin42^circ sec48^circ + cos42^circ cosec48^circ`
⇒ `sin42^circ sec(90^circ - 42^circ) + cos42^circ cosec(90^circ - 42^circ)`
⇒ `sin42^circ . cosec42^circ + cos42^circ sec42^circ`
⇒ `sin42^circ . 1/sin42^circ + cos42^circ 1/cos42^circ`
⇒ 1 + 1 = 2
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