prove this identity =>
. [ 1+ tan²A / 1+cot²A] = [ 1-tan A/1-cot A] = tan² A
dhathri123:
i am getting the second one as -tanA
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Answered by
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[ 1+ tan²A / 1+cot²A] = [ 1+ tan²A / 1+1/tan²A]
=1+tan²A/tan²A+1/tan²A
=1+tan²A. tan²A/tan²A+1
=tan²A. --------(1)
[ 1-tanA/1-cotA]² = [ 1²-tan²A/1²-1/tan²A]
=1-tan²A/tan²A-1/tan²A
=1-tan²A. tan²A/tan²A-1
=tan²A. ---------(2)
eq. (1)&(2) are equal...
[ 1+ tan²A / 1+cot²A] = [ 1-tan A/1-cot A]² = tan² A
hence proved..
(*bt in this question if the squares of tan and cot will be there then only this can be proved*)
Hope this helped you..
=1+tan²A/tan²A+1/tan²A
=1+tan²A. tan²A/tan²A+1
=tan²A. --------(1)
[ 1-tanA/1-cotA]² = [ 1²-tan²A/1²-1/tan²A]
=1-tan²A/tan²A-1/tan²A
=1-tan²A. tan²A/tan²A-1
=tan²A. ---------(2)
eq. (1)&(2) are equal...
[ 1+ tan²A / 1+cot²A] = [ 1-tan A/1-cot A]² = tan² A
hence proved..
(*bt in this question if the squares of tan and cot will be there then only this can be proved*)
Hope this helped you..
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