can y'all pls solve this??? ASAP!!!!
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priyanshupatel83:
Answer will be 1
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Answered by
1
hey, ur ans is 1. becz...
= cos2(90° -B )/ sin2 B
= sin2 B / sin2 B
= 1...
= cos2(90° -B )/ sin2 B
= sin2 B / sin2 B
= 1...
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Answered by
1
If B is an acute angle , then
![\frac{ { \cos}^{2}(90 - B) }{ { \sin }^{2}B } \\ \\ = > \frac{ {[ \cos(90 - B) ]}^{2} }{ { \sin}^{2}B } \\ \\ = > \frac{ [{ \sin(B) }]^{2} }{ { \sin}^{2} B } \\ \\ = > \frac{ { \sin }^{2} B }{ { \sin }^{2} B } \\ \\ = > 1 \\ \\ As \: we \: know \: that :\: \cos(90 - \alpha ) = \sin \alpha](https://tex.z-dn.net/?f=+%5Cfrac%7B+%7B+%5Ccos%7D%5E%7B2%7D%2890+-+B%29+%7D%7B+%7B+%5Csin+%7D%5E%7B2%7DB+%7D+%5C%5C+%5C%5C+%3D+%26gt%3B+%5Cfrac%7B+%7B%5B+%5Ccos%2890+-+B%29+%5D%7D%5E%7B2%7D+%7D%7B+%7B+%5Csin%7D%5E%7B2%7DB+%7D+%5C%5C+%5C%5C+%3D+%26gt%3B+%5Cfrac%7B+%5B%7B+%5Csin%28B%29+%7D%5D%5E%7B2%7D+%7D%7B+%7B+%5Csin%7D%5E%7B2%7D+B+%7D+%5C%5C+%5C%5C+%3D+%26gt%3B+%5Cfrac%7B+%7B+%5Csin+%7D%5E%7B2%7D+B+%7D%7B+%7B+%5Csin+%7D%5E%7B2%7D+B+%7D+%5C%5C+%5C%5C+%3D+%26gt%3B+1+%5C%5C+%5C%5C+As+%5C%3A+we+%5C%3A+know+%5C%3A+that+%3A%5C%3A+%5Ccos%2890+-+%5Calpha+%29+%3D+%5Csin+%5Calpha+ "\frac{ { \cos}^{2}(90 - B) }{ { \sin }^{2}B } \\ \\ = > \frac{ {[ \cos(90 - B) ]}^{2} }{ { \sin}^{2}B } \\ \\ = > \frac{ [{ \sin(B) }]^{2} }{ { \sin}^{2} B } \\ \\ = > \frac{ { \sin }^{2} B }{ { \sin }^{2} B } \\ \\ = > 1 \\ \\ As \: we \: know \: that :\: \cos(90 - \alpha ) = \sin \alpha")
sorry buddy i think ur mathematical concepts are weak
but i done it correct so don't worry
hi
I am aisha
^_^
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