Prove that: cos2A + tan2A - 1 / sin2A = tan2A
Answers
The equality \frac{cos^2a+tan^2a-1}{sin^2a}=tan^2a
sin
2
a
cos
2
a+tan
2
a−1
=tan
2
a is proved
Step-by-step explanation:
Given that \frac{cos^2a+tan^2a-1}{sin^2a}=tan^2a
sin
2
a
cos
2
a+tan
2
a−1
=tan
2
a
To prove the given equality :
Taking LHS
\frac{cos^2a+tan^2a-1}{sin^2a}
sin
2
a
cos
2
a+tan
2
a−1
=\frac{cos^2a+tan^2a-(cos^2a+sin^2a)}{sin^2a}=
sin
2
a
cos
2
a+tan
2
a−(cos
2
a+sin
2
a)
( by using the identity cos^2x+sin^2x=1cos
2
x+sin
2
x=1 here x=a )
=\frac{cos^2a+tan^2a-cos^2a-sin^2a}{sin^2a}=
sin
2
a
cos
2
a+tan
2
a−cos
2
a−sin
2
a
=\frac{tan^2a-sin^2a}{sin^2a}=
sin
2
a
tan
2
a−sin
2
a
=\frac{tan^2a}{sin^2a}-\frac{sin^2a}{sin^2a}=
sin
2
a
tan
2
a
−
sin
2
a
sin
2
a
=\frac{tan^2a}{sin^2a}-1=
sin
2
a
tan
2
a
−1
=\frac{\frac{sin^2a}{cos^2a}}{sin^2a}-1=
sin
2
a
cos
2
a
sin
2
a
−1 ( we know tanx=\frac{sinx}{cosx}tanx=
cosx
sinx
)
=(\frac{sin^2a}{cos^2a}\times \frac{1}{sin^2a})-1=(
cos
2
a
sin
2
a
×
sin
2
a
1
)−1
=\frac{1}{cos^2a}-1=
cos
2
a
1
−1
=sec^2a-1=sec
2
a−1 ( by using secx=\frac{1}{cosx}secx=
cosx
1
)
=tan^2a=RHS=tan
2
a=RHS ( by using the identity sec^2x-tan^2x=1sec
2
x−tan
2
x=1 )
Therefore LHS=RHS
The equality \frac{cos^2a+tan^2a-1}{sin^2a}=tan^2a
sin
2
a
cos
2
a+tan
2
a−1
=tan
2
a is proved