Question

trigonometry question

Answer 1

hey !!!

(tan¢ + cot¢ )²- cosec²¢ - sec²¢ = 0

from LHS

(tan¢ + cot¢ )² - cosec²¢ - sec²¢

=> tan²¢ + cot²¢ + 2tan¢ × cot¢ - cosec²¢ - sec²¢

=> tan²¢ - sec²¢ - cosec²¢ + cot²¢ + 2tan¢ × cot¢

=>-( sec²¢ - tan²¢ ) -( cosec²¢ - cot²¢ ) + 2tan¢ × cot¢

=> -1 - 1 + 2

=.> - 2 + 2

= > 0 RHS prooved

hope it helps !!!

₹Rajukumar111

Answer 2

Question:

Show that (tanθ + cotθ)² - cosec²θ - sec²θ = 0

Step-by-step explanation:

L.H.S. : (tanθ + cotθ)² - cosec²θ - sec²θ

• Identity : (a + b)² = a² + b² + 2ab Here, a = tanθ, b = cotθ

→ [ (tanθ)² + (cotθ)² + 2(tanθ)(cotθ) ] - cosec ²θ - sec²θ

→ [ tan²θ + cot²θ + 2.tanθ.cotθ ] - cosec²θ - sec²θ

Opening the bracket.

→ tan²θ + cot²θ + 2.tanθ.cotθ - cosec²θ - sec²θ

• Identity : 1 + tan²θ = sec²θ

From this, we get [ tan²θ = sec²θ - 1 ]

• Identity : 1 + cot²θ = cosec²θ

From this, we get [ cot²θ = cosec²θ - 1 ]

→ (sec²θ - 1) + (cosec²θ - 1) + 2.tanθ.cotθ - cosec²θ - sec²θ

Opening the brackets.

→ sec²θ - 1 + cosec²θ - 1 + 2.tanθ.cotθ - cosec²θ - sec²θ

Rearranging the terms.

→ sec²θ - sec²θ + cosec²θ - cosec²θ - 1 - 1 + 2.tanθ.cotθ

→ - 1 - 1 + 2.tanθ.cotθ

• Identity : cotθ = 1/tanθ

→ - 1 - 1 + 2.tanθ.(1/tanθ)

→ - 1 - 1 + 2

→ - (1 + 1) + 2

→ - 2 + 2

→ 0

= R.H.S.

Hence, proved !!