if cosA=2/5, then find the value of 4+4tan^2A
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Answered by
167
cosA=2/5
sinaA=√{1-(2/5)^2}
=√1-4/25
=√21/25=(√21)/5
tanA=sinA/cosA
=(√21)/5÷2/5
=(√21)/2
LHS
4+4tan^2A
=4+4*{(√21)/2}^2
=4+4*21/4
=4+21
=25
sinaA=√{1-(2/5)^2}
=√1-4/25
=√21/25=(√21)/5
tanA=sinA/cosA
=(√21)/5÷2/5
=(√21)/2
LHS
4+4tan^2A
=4+4*{(√21)/2}^2
=4+4*21/4
=4+21
=25
Xahil:
kuch samaj nhi aaya
Very nice explaination
Answered by
137
hope it help mrk it as brainliest...
there is another method of solving it by mading a triangle but i did it with other method....
there is another method of solving it by mading a triangle but i did it with other method....
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