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\begin{array}{l}{A+B=90^{\circ}, \text { prove that }} \\ {\frac{\tan A \tan B+\tan A \cot B}{\sin A \sec B}-\frac{\sin ^{2} B}{\cos ^{2} A}=\tan A}\end{array} |
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Answer» Tan 1 . tan 2 .tan 3 . .......tan 88 . tan 89=sin1/cos1 .sin 2/cos 2 . ...sin 45/cos45..sin 88/cos 88 . sin 89/ cos 89=sin(90-89)/cos1 . sin(90-88)/cos2...........sin(90-45)/cos 45.........sin(90-1)/cos89=cos 89/cos1 .cos 88/cos 2......cos 45/cos 45 ..........cos 1/ cos 89 all the values will get cancelled......and the result will be 1 |
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