Question

prove that 1-tan raise to the power 4 theta divided by 1+ tan raise to the power 4 theta= cos theta+ sin theta divided by cos theta - sin theta​

Answer 1

$\dfrac{1-\tan^4 \theta}{1+\tan^4 \theta} =\dfrac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}$, proved.

Step-by-step explanation:

To prove that, $\dfrac{1-\tan^4 \theta}{1+\tan^4 \theta} =\dfrac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}$.

L.H.S.$=\dfrac{1-\tan^4 \theta}{1+\tan^4 \theta}$

= $\dfrac{1-\dfrac{\sin^4 \theta}{\cos^4 \theta}}{1+\dfrac{\sin^4 \theta}{\cos^4 \theta}}$

= $\dfrac{\cos^4 \theta-\sin^4 \theta}{\cos^4 \theta+\sin^4 \theta}$

= $\dfrac{(\cos^2 \theta)^2-(\sin^2 \theta)^2}{(\cos^2 \theta)^2+(\sin^2 \theta)^2}$

= $\dfrac{(\cos^2 \theta+\sin^2 \theta)(\cos^2 \theta-\sin^2 \theta)}{(\cos^2 \theta)^2+(\sin^2 \theta)^2}$

Using the algebraic identity,

$a^{2}-b^{2}=(a+b)(a-b)$

= $\dfrac{(1)(\cos^2 \theta-\sin^2 \theta)}{(\cos^2 \theta+\sin^2 \theta)^2-2\cos^2 \theta\sin^2 \theta}$

Using the trigonometric identity,

$\sin^2 A+\cos^2 A = 1$

= $\dfrac{(\cos \theta+\sin \theta)(\cos \theta-\sin \theta)}{(1)^2-2\cos^2 \theta.\sin^2 \theta}$

= $\dfrac{1+\sin 2\theta}{\cos 2\theta}$

R.H.S. = $\dfrac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}$

= $\dfrac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}\times \dfrac{\cos \theta+\sin \theta}{\cos \theta+\sin \theta}$

= $\dfrac{(\cos \theta+\sin \theta)^2}{\cos^2 \theta-\sin^2 \theta}$

$=\dfrac{\cos^2 \theta+\sin^2 \theta+2\sin \theta\cos \theta}{\cos^2 \theta-\sin^2 \theta}$

Using the trigonometric identity,

$\sin 2A=2\sin A\cos A$ and

$\cos 2A=\cos^2 A-\sin^2 A$

$= \dfrac{1+\sin 2\theta}{\cos 2\theta}$

Hence, $\dfrac{1-\tan^4 \theta}{1+\tan^4 \theta} =\dfrac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}$, proved.

Answer 2

Answer:

\dfrac{1-\tan^4 \theta}{1+\tan^4 \theta} =\dfrac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}

1+tan

4

θ

1−tan

4

θ

=

cosθ−sinθ

cosθ+sinθ

, proved.

Step-by-step explanation:

To prove that, \dfrac{1-\tan^4 \theta}{1+\tan^4 \theta} =\dfrac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}

1+tan

4

θ

1−tan

4

θ

=

cosθ−sinθ

cosθ+sinθ

.

L.H.S.=\dfrac{1-\tan^4 \theta}{1+\tan^4 \theta}=

1+tan

4

θ

1−tan

4

θ

= \dfrac{1-\dfrac{\sin^4 \theta}{\cos^4 \theta}}{1+\dfrac{\sin^4 \theta}{\cos^4 \theta}}

1+

cos

4

θ

sin

4

θ

1−

cos

4

θ

sin

4

θ

= \dfrac{\cos^4 \theta-\sin^4 \theta}{\cos^4 \theta+\sin^4 \theta}

cos

4

θ+sin

4

θ

cos

4

θ−sin

4

θ

= \dfrac{(\cos^2 \theta)^2-(\sin^2 \theta)^2}{(\cos^2 \theta)^2+(\sin^2 \theta)^2}

(cos

2

θ)

2

+(sin

2

θ)

2

(cos

2

θ)

2

−(sin

2

θ)

2

= \dfrac{(\cos^2 \theta+\sin^2 \theta)(\cos^2 \theta-\sin^2 \theta)}{(\cos^2 \theta)^2+(\sin^2 \theta)^2}

(cos

2

θ)

2

+(sin

2

θ)

2

(cos

2

θ+sin

2

θ)(cos

2

θ−sin

2

θ)

Using the algebraic identity,

a^{2}-b^{2}=(a+b)(a-b)a

2

−b

2

=(a+b)(a−b)

= \dfrac{(1)(\cos^2 \theta-\sin^2 \theta)}{(\cos^2 \theta+\sin^2 \theta)^2-2\cos^2 \theta\sin^2 \theta}

(cos

2

θ+sin

2

θ)

2

−2cos

2

θsin

2

θ

(1)(cos

2

θ−sin

2

θ)

Using the trigonometric identity,

\sin^2 A+\cos^2 A = 1sin

2

A+cos

2

A=1

= \dfrac{(\cos \theta+\sin \theta)(\cos \theta-\sin \theta)}{(1)^2-2\cos^2 \theta.\sin^2 \theta}

(1)

2

−2cos

2

θ.sin

2

θ

(cosθ+sinθ)(cosθ−sinθ)

= \dfrac{1+\sin 2\theta}{\cos 2\theta}

cos2θ

1+sin2θ

R.H.S. = \dfrac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}

cosθ−sinθ

cosθ+sinθ

= \dfrac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}\times \dfrac{\cos \theta+\sin \theta}{\cos \theta+\sin \theta}

cosθ−sinθ

cosθ+sinθ

×

cosθ+sinθ

cosθ+sinθ

= \dfrac{(\cos \theta+\sin \theta)^2}{\cos^2 \theta-\sin^2 \theta}

cos

2

θ−sin

2

θ

(cosθ+sinθ)

2

=\dfrac{\cos^2 \theta+\sin^2 \theta+2\sin \theta\cos \theta}{\cos^2 \theta-\sin^2 \theta}=

cos

2

θ−sin

2

θ

cos

2

θ+sin

2

θ+2sinθcosθ

Using the trigonometric identity,

\sin 2A=2\sin A\cos Asin2A=2sinAcosA and

\cos 2A=\cos^2 A-\sin^2 Acos2A=cos

2

A−sin

2

A

= \dfrac{1+\sin 2\theta}{\cos 2\theta}=

cos2θ

1+sin2θ

Hence, \dfrac{1-\tan^4 \theta}{1+\tan^4 \theta} =\dfrac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}

1+tan

4

θ

1−tan

4

θ

=

cosθ−sinθ

cosθ+sinθ

, proved.