Question

prove that (tan theta/sec theta-1)-(sin theta /1+cos theta)=2cot theta
​

Answer 1

GIVEN :

The equation is \frac{tan\theta}{sec\theta-1}-\frac{sin\theta}{1+cos\theta}=2cot\theta

secθ−1

tanθ

−

1+cosθ

sinθ

=2cotθ

TO PROVE :

The given equation \frac{tan\theta}{sec\theta-1}-\frac{sin\theta}{1+cos\theta}=2cot\theta

secθ−1

tanθ

−

1+cosθ

sinθ

=2cotθ is true

SOLUTION :

From the given equation is true

That is prove that LHS = RHS

Now taking LHS

\frac{tan\theta}{sec\theta-1}-\frac{sin\theta}{1+cos\theta}

secθ−1

tanθ

−

1+cosθ

sinθ

By using the trignometric identity :

secx=\frac{1}{cosx}secx=

cosx

1

=\frac{\frac{sin\theta}{cos\theta}}{\frac{1}{cos\theta}-1}-\frac{sin\theta}{1+cos\theta}=

cosθ

1

−1

cosθ

sinθ

−

1+cosθ

sinθ

=\frac{\frac{sin\theta}{cos\theta}}{\frac{1-cos\theta}{cos\theta}}-\frac{sin\theta}{1+cos\theta}=

cosθ

1−cosθ

cosθ

sinθ

−

1+cosθ

sinθ

=\frac{sin\theta}{cos\theta}}\times (\frac{cos\theta}{1-cos\theta})-\frac{sin\theta}{1+cos\theta}

=\frac{sin\theta}{1-cos\theta}-\frac{sin\theta}{1+cos\theta}=

1−cosθ

sinθ

−

1+cosθ

sinθ

=\frac{sin\theta(1+cos\theta)-sin\theta(1-cos\theta)}{(1-cos\theta)(1+cos\theta)}=

(1−cosθ)(1+cosθ)

sinθ(1+cosθ)−sinθ(1−cosθ)

By using the algebraic identity :

(a+b)(a-b)-a^2-b^2(a+b)(a−b)−a

2

−b

2

=\frac{sin\theta+sin\theta cos\theta-sin\theta+sin\theta cos\theta}{1^2-cos^2\theta}=

1

2

−cos

2

θ

sinθ+sinθcosθ−sinθ+sinθcosθ

Adding the like terms

=\frac{2sin\theta cos\theta}{sin^2\theta}=

sin

2

θ

2sinθcosθ

=2(\frac{cos\theta}{sin\theta})=2(

sinθ

cosθ

)

By using the trignometric identity :

cotx=\frac{cos}{sinx}cotx=

sinx

cos

=2cot\theta=2cotθ =RHS

∴ LHS = RHS

∴ the equation \frac{tan\theta}{sec\theta-1}-\frac{sin\theta}{1+cos\theta}=2cot\theta

secθ−1

tanθ

−

1+cosθ

sinθ

=2cotθ is true

Hence proved.