If 3cotA = 2 show tat sec²A = 1 + tan²A with the help of a right triangle?.
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Step-by-step explanation:
∠B=90
o
cotA=
B
A
=
3
4
Let AB=4x,BC=3x.
AC
2
=AB+BC
2
AC
2
=16k
2
+9x
2
AC
2
=25x
2
AC=5x
Now, tanA=
4
3
LHS :
1+tan
2
A
1−tan
2
A
=
1+
16
9
1−
16
9
=
25
7
RHS : cos
2
A−sin
2
A=(
5
4
)
2
−(
5
3
)
2
=
25
16
−
25
9
=
25
7
⇒LHS=RHS
∴
1+tan
2
A
1−tan
2
A
=cos
2
A−sin
2
A
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