Question

If sec theta+tan theta=m, then sec^4 theta-tan^4 theta-2sec theta×tan theta=?

Answer 1

$\sec^4 \theta-\tan^4 \theta-2\sec \theta\times\tan \theta=\dfrac{1}{m^2}$

Step-by-step explanation:

We have,

$\sec \theta+\tan \theta=m$

To find, $\sec^4 \theta-\tan^4 \theta-2\sec \theta\times\tan \theta=?$

∴ $\sec^4 \theta-\tan^4 \theta-2\sec \theta\times\tan \theta$

$=(\sec^2 \theta)^2-(\tan^2 \theta)^2-2\sec \theta\times\tan \theta$

$=(\sec^2 \theta +\tan^2 \theta)(\sec^2 \theta -\tan^2 \theta)-2\sec \theta\times\tan \theta$

[ ∵ $a^{2} -b^{2} =(a+b)(a-b)$]

$=(\sec^2 \theta +\tan^2 \theta)(1)-2\sec \theta\times\tan \theta$

[ ∵ $\sec^2 \theta -\tan^2 \theta=1$]

$=\sec^2 \theta +\tan^2 \theta-2\sec \theta\times\tan \theta$

$=(\sec \theta-\tan \theta)^2$     ......(1)

[ ∵ $(a-b)^{2} =a^{2} +b^{2} -2ab$]

Given by question,

$\sec \theta+\tan \theta=m$

Rationalising numerator and denominator, we get

$\dfrac{(\sec \theta+\tan \theta)(\sec \theta-\tan \theta)}{\sec \theta-\tan \theta} =m$

⇒ $\dfrac{\sec^2 \theta-\tan^2 \theta}{\sec \theta-\tan \theta} =m$

⇒ $\dfrac{1}{\sec \theta-\tan \theta} =m$

⇒ $\sec \theta-\tan \theta}=\dfrac{1}{m}$

Put $\sec \theta-\tan \theta}=\dfrac{1}{m}$ in (1), we get

$\sec^4 \theta-\tan^4 \theta-2\sec \theta\times\tan \theta=(\dfrac{1}{m})^2=\dfrac{1}{m^2}$

∴ $\sec^4 \theta-\tan^4 \theta-2\sec \theta\times\tan \theta=\dfrac{1}{m^2}$

Answer 2

Answer:

sec

4

θ−tan

4

θ−2secθ×tanθ=

m

2

1

Step-by-step explanation:

We have,

\sec \theta+\tan \theta=msecθ+tanθ=m

To find, \sec^4 \theta-\tan^4 \theta-2\sec \theta\times\tan \theta=?sec

4

θ−tan

4

θ−2secθ×tanθ=?

∴ \sec^4 \theta-\tan^4 \theta-2\sec \theta\times\tan \thetasec

4

θ−tan

4

θ−2secθ×tanθ

=(\sec^2 \theta)^2-(\tan^2 \theta)^2-2\sec \theta\times\tan \theta=(sec

2

θ)

2

−(tan

2

θ)

2

−2secθ×tanθ

=(\sec^2 \theta +\tan^2 \theta)(\sec^2 \theta -\tan^2 \theta)-2\sec \theta\times\tan \theta=(sec

2

θ+tan

2

θ)(sec

2

θ−tan

2

θ)−2secθ×tanθ

[ ∵ a^{2} -b^{2} =(a+b)(a-b)a

2

−b

2

=(a+b)(a−b)

=(\sec^2 \theta +\tan^2 \theta)(1)-2\sec \theta\times\tan \theta=(sec

2

θ+tan

2

θ)(1)−2secθ×tanθ

[ ∵ \sec^2 \theta -\tan^2 \theta=1sec

2

θ−tan

2

θ=1

=\sec^2 \theta +\tan^2 \theta-2\sec \theta\times\tan \theta=sec

2

θ+tan

2

θ−2secθ×tanθ

=(\sec \theta-\tan \theta)^2=(secθ−tanθ)

2

......(1)

[ ∵ (a-b)^{2} =a^{2} +b^{2} -2ab(a−b)

2

=a

2

+b

2

−2ab

Given by question,

\sec \theta+\tan \theta=msecθ+tanθ=m

Rationalising numerator and denominator, we get

\dfrac{(\sec \theta+\tan \theta)(\sec \theta-\tan \theta)}{\sec \theta-\tan \theta} =m

secθ−tanθ

(secθ+tanθ)(secθ−tanθ)

=m

⇒ \dfrac{\sec^2 \theta-\tan^2 \theta}{\sec \theta-\tan \theta} =m

secθ−tanθ

sec

2

θ−tan

2

θ

=m

⇒ \dfrac{1}{\sec \theta-\tan \theta} =m

secθ−tanθ

1

=m

⇒ \sec \theta-\tan \theta}=\dfrac{1}{m}

Put \sec \theta-\tan \theta}=\dfrac{1}{m} in (1), we get

\sec^4 \theta-\tan^4 \theta-2\sec \theta\times\tan \theta=(\dfrac{1}{m})^2=\dfrac{1}{m^2}sec

4

θ−tan

4

θ−2secθ×tanθ=(

m

1

)

2

=

m

2

1

∴ \sec^4 \theta-\tan^4 \theta-2\sec \theta\times\tan \theta=\dfrac{1}{m^2}sec

4

θ−tan

4

θ−2secθ×tanθ=

m

2

1