Show that: (i) sin 20°. cos 40° + c...
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Show that: (i) sin 20°. cos 40° + cos 20°. sin 40° = 1312 (ii) cos 100°. cos 40° + sin 100°. sin 40º = 1/2

JEE/Engineering Exams
Maths
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( 3-4 ) (i) ( sin 20^{circ}-cos 40^{circ}+cos 20 . sin 40^{circ} ) ( sin (90-70) ) (i) ( sin 20^{circ} cdot cos 40+cos 20 cdot sin 40 ) ( sin A cos B+cos A sin B=sin (A+B) ) ( sin (20+40)=sin 60^{circ}=1312 ) (ii) ( cos 100^{circ} cdot cos 40^{circ}+8 operatorname{in} 100 cdot sin 40^{circ} ) ( cos A cos B+sin A sin B=cos (A-B) ) ( cos (100-40)=cos 60^{circ}=42 . ) ( B-3 quad tan theta+sec theta=2 / 3 ) ( sqrt{x}+6 x^{2}+sec ^{2} theta+x tan theta sec theta=pi / 9 ) ( (tan theta+sec theta)(tan theta-sec theta)=frac{2}{3}(tan theta-sec theta) ) ( tan ^{2} theta-sec ^{2} theta=2 sin (tan theta-sec theta) ) ( 3 / 2=tan theta-sec theta ) ( cos (1)-(2) ) ( 2 sec theta=2 / 3-3 / 2=frac{4-9}{6} ) ( 1^{5} sec theta=-5 / 12 )
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