Prove that: (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan? A + cot? A
Answers
EXPLANATION.
⇒ (sin(A) + cosec(A))² + (cos(A) + sec(A))² = 7 + tan²A + cot²A.
As we know that,
From L.H.S we get,
⇒ (sin(A) + cosec(A))² + (cos(A) + sec(A))²
As we know that,
Formula of :
⇒ (x + y)² = x² + y² + 2xy.
Using this formula in the equation, we get.
⇒ sin²A + cosec²A + 2.sin(A).cosec(A) + cos²A + sec²A + 2.cos(A).sec(A).
⇒ sin²A + cosec²A + 2.sin(A) x 1/sin(A) + cos²A + sec²A + 2.cos(A) x 1/cos(A).
⇒ sin²A + cosec²A + 2 + cos²A + sec²A + 2.
⇒ sin²A + cos²A + cosec²A + sec²A + 4.
⇒ 1 + (1 + cot²A) + (1 + tan²A) + 4.
⇒ 1 + 1 + cot²A + 1 + tan²A + 4.
⇒ 7 + tan²A + cot²A.
Hence Proved.
MORE INFORMATION.
(1) = sin²θ + cos²θ = 1.
(2) = 1 + tan²θ = sec²θ.
(3) = 1 + cot²θ = cosec²θ.
(4) = sin2θ = 2sinθ.cosθ = 2tanθ/1 + tan²θ.
(5) = cos2θ = cos²θ - sin²θ = 2cos²θ - 1 = 1 - 2sin²θ = 1 - tan²θ/1 + tan²θ.
(6) = tan2θ = 2tanθ/1 - tan²θ.
(7) = sin3θ = 3sinθ - 4sin³θ.
(8) = cos3θ = 4cos³θ - 3cosθ.
(9) = tan3θ = 3tanθ - tan³θ/1 - 3tan²θ.
SOLUTION GIVEN IN ATTACHMENT.
