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