Verified answer

Solution:

Given that:

[tex]\tt\longrightarrow l=m\:sec\:\theta+n\:tan\:\theta[/tex]

Squaring both sides, we get:

[tex]\tt\longrightarrow l^2=(m\:sec\:\theta+n\:tan\:\theta)^2[/tex]

[tex]\tt\longrightarrow l^2=m^2sec^2\theta+n^2tan^2\theta+2\:mn\:sec\:\theta\:tan\:\theta-(i)[/tex]

Again:

[tex]\tt\longrightarrow k=n\:sec\:\theta+m\:tan\:\theta[/tex]

Squaring both sides, we get:

[tex]\tt\longrightarrow k^2=n^2sec^2\theta+m^2tan^2\theta+2\:mn\:sec\:\theta\:tan\:\theta-(ii)[/tex]

Subtracting (ii) from (i), we get:

[tex]\tt\longrightarrow l^2-k^2=(m^2sec^2\theta+n^2tan^2\theta+2\:mn\:sec\:\theta\:tan\:\theta)-(n^2sec^2\theta+m^2tan^2\theta+2\:mn\:sec\:\theta\:tan\:\theta)[/tex]

[tex]\tt\longrightarrow l^2-k^2=(m^2sec^2\theta-m^2tan^2\theta)+(n^2tan^2\theta-n^2sec^2\theta)+(2\:mn\:sec\:\theta\:tan\:\theta-2\:mn\:sec\:\theta\:tan\:\theta)[/tex]

[tex]\tt\longrightarrow l^2-k^2=m^2(sec^2\theta-tan^2\theta)-n^2(sec^2\theta-tan^2\theta)[/tex]

We know that:

[tex]\bigstar\:\:\underline{\boxed{\tt sec^x-tan^2x=1}}[/tex]

Using this result, we get:

[tex]\tt\longrightarrow l^2-k^2=m^2\cdot 1-n^2\cdot 1[/tex]

[tex]\tt\longrightarrow l^2-k^2=m^2-n^2[/tex]

Which is our required answer.

Answer:

[tex]\tt\hookrightarrow l^2-k^2=m^2-n^2[/tex]

Learn More:

1. Relationship between sides and T-Ratios.

• sin θ = Height/Hypotenuse
• cos θ = Base/Hypotenuse
• tan θ = Height/Base
• cot θ = Base/Height
• sec θ = Hypotenuse/Base
• cosec θ = Hypotenuse/Height

2. Square formulae.

• sin²θ + cos²θ = 1
• cosec²θ - cot²θ = 1
• sec²θ - tan²θ = 1

3. Reciprocal Relationship.

• sin θ = 1/cosec θ
• cos θ = 1/sec θ
• tan θ = 1/cot θ
• cosec θ = 1/sin θ
• sec θ = 1/cos θ
• tan θ = 1/cot θ

4. Cofunction identities.

• sin(90° - θ) = cos θ
• cos(90° - θ) = sin θ
• cosec(90° - θ) = sec θ
• sec(90° - θ) = cosec θ
• tan(90° - θ) = cot θ
• cot(90° - θ) = tan θ

5. Even odd identities.

• sin -θ = -sin θ
• cos -θ = cos θ
• tan -θ = -tan θ