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\frac{\sin 50^{\circ}}{\cos 40^{\circ}}+\frac{\csc 40^{\circ}}{\sec 50^{\circ}}-4 \cos 50^{\circ} \cdot \csc 40^{\circ} |
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Answer» Taking L.H.S; sin 50/ cos 40 + cosec 40/ sec 50 - 4 cos 50 * cosec 40= sin 50/ cos 40 + cos 50/ sin 40 - 4 cos 50/sin 40[ since cosec A= 1/sinA and sec A= 1/cosA.]=cos(90-50)/cos 40+ sin(90-50)/sin 40 - 4 sin(90-50)/sin 40[since cos A= sin (90-A) and sin A= cos (90-A)] = cos 40/cos 40 + sin 40/sin 40 - 4 sin 40/sin 40=1+1-4*1=1+1-4=2-4=-2=R.H.S Hence,proved. sin(90–40)/cos40 + cosec(90–50)/sec50 –4cos(90–40).1/sin40= =cos40/cos40+sec50/sec50–4sin40.1/sin40= 1+1-4=-2 |
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