solve this question of trigonometry
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sin(A+B)=sinAcosB+cosAsinB
so, sin20cos70+cos20sin70 = sin(20+70) = sin90 = 1
cosec a = 1/sina and sec a= 1/cosa
so, sin23cosec23+cos23sec23 = sin23 x (1/sin23) + cos23 x (1/cos23)
= 1+1 = 2
therefore, (sin20cos70 + cos20sin70)/(sin23cosec23+cos23sec23) = 1/2
so, sin20cos70+cos20sin70 = sin(20+70) = sin90 = 1
cosec a = 1/sina and sec a= 1/cosa
so, sin23cosec23+cos23sec23 = sin23 x (1/sin23) + cos23 x (1/cos23)
= 1+1 = 2
therefore, (sin20cos70 + cos20sin70)/(sin23cosec23+cos23sec23) = 1/2
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