Question

Prove that (Sin theta+cosec theta)^2 + (cos theta+sec theta)^2=7+tan^2theta+cot^2theta

Answer 1

hence proved
7+ tan^2x+cot^2x

Answer 2

Answer:

( sin θ + cosec θ )² + ( cos θ + sec θ )² = 7 + tan² θ + cot² θ

Step-by-step-explanation:

We have to prove the trigonometric equation

( sin θ + cosec θ )² + ( cos θ + sec θ )² = 7 + tan² θ + cot² θ

∴ LHS = ( sin θ + cosec θ )² + ( cos θ + sec θ )²

We know that,

( a + b )² = a² + 2ab + b²

⇒ LHS = sin² θ + cosec² θ + 2 sin θ cosec θ + cos² θ + sec² θ + 2 cos θ sec θ

We know that,

• cosec θ = 1 / sin θ

• sec θ = 1 / cos θ

⇒ LHS = sin² θ + cosec² θ + 2 sin θ * 1 / sin θ + cos² θ + sec² θ + 2 cos θ * 1 / cos θ

⇒ LHS = sin² θ + cosec² θ + 2 + cos² θ + sec² θ + 2

⇒ LHS = sin² θ + cos² θ + 2 + 2 + cosec² θ + sec² θ

⇒ LHS = 1 + 4 + cosec² θ + sec² θ

We know that,

cosec² θ = 1 + cot² θ

⇒ LHS = 5 + 1 + cot² θ + sec² θ

⇒ LHS = 6 + cot² θ + sec² θ

We know that,

sec² θ = 1 + tan² θ

⇒ LHS = 6 + cot² θ + 1 + tan² θ

⇒ LHS = 6 + 1 + tan² θ + cot² θ

⇒ LHS = 7 + tan² θ + cot² θ

RHS = 7 + tan² θ + cot² θ

∴ LHS = RHS

Hence proved!