# Prove the following trigonometric identities (a) sec^6A - tan^6A=1 + 3 tan^2 A * Sec^2 A (b) cos^4A - sin^4A = 2 cos^2 a - 1 (a) sec^6A - tan^6A=1 + 3 tan^2 A * Sec^2 A (b) cos^4A - sin^4A = 2...

Prove the following trigonometric identities

(a) sec^6A - tan^6A=1 + 3 tan^2 A * Sec^2 A

(b) cos^4A - sin^4A = 2 cos^2 a - 1

(a) sec^6A - tan^6A=1 + 3 tan^2 A * Sec^2 A

(b) cos^4A - sin^4A = 2 cos^2 a - 1

(c) sin A (1+ tan A) + cos A ( 1 + cot A) = sec A+ cosec A

I want answers for these questions because I have Exam in this month.

So please...........

Asked on by chetan3618

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lochana2500 | Student, Undergraduate | (Level 1) Valedictorian

Posted on

a). sec⁶A - tan⁶A = (sec²A)³ - (tan²A)³

⇒ use x³ - y³ = (x - y) (x² + y² + xy)

= (sec²A - tan²A) [ (sec²A)² + (tan²A)² + tan²A.sec²A ]

⇒ use sec²A - tan²A = 1

= [ (sec²A)² + (tan²A)² + tan²A.sec²A ]

⇒ use x² + y² = (x-y)² + 2xy

= (sec²A - tan²A)² + 2tan²A.sec²A + tan²A.sec²A

= 1 + 3tan²A.sec²A

L:H:S ≡ R:H:S

b). cos⁴A - sin⁴A = (cos²A)² - (sin²A)²

= (cos²A + sin²A) (cos²A - sin²A)

⇒ use cos²A + sin²A = 1

= (cos²A - sin²A)

⇒ use sin²A = 1 - cos²A

= cos²A -1 + cos²A

= 2cos²A -1

L:H:S ≡ R:H:S

c).

sin A (1+ tan A) + cos A ( 1 + cot A) = sinA(1+tanA)+cos(1+cotA)

= sinA+sin²A/cosA + cosA +cos²A/sinA

= sinA +(1-cos²A)/cosA +cosA +(1-sin²A)/sinA

= sinA+1/cosA -cosA +cosA +1/sinA - sinA

= 1/sinA +1/cosA

= cosecA + secA

L:H:S ≡ R:H:S

Top Answer

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To prove

sec^6-tan^6A = 1+3tan^2A*sec^2A.

Put sec^2A = x and tan^2A = y. Then sec^2A-tan^2A= 1 is also an identity. So x-y = 1. Using these, we get:

(x^3-y^3) = (x-y)(x^2+xy+y^2) is andentity.

sec^6A - tan^6A = 1*(sec^4A +sec^2A*tan^2A+tan^4A) = {(1+tan^2)sec^2A + sec^2A*tan^2A+ } =

(sec^2A+tan^2Asec^2A +sec^2Atan^2A+(sec^2A-1)tan^2A) = (3sec^2Atan^2A +sec^A-tan^2A)

= 3tan^2Asec^2A+1, as sec^A-tan^2A = 1.

Threrefore,  sec^6A -tan^6A = 1+3tan^2A*sec^2A.

b)

We use a^2-b^2 = (a-b) (a+b) and  sin^x+ cos^2x = 1

LHS = (cos^4A-sin^4A = (cos^2A-sin^2A)(cos^2A+sin^2A) ,

= cos^2A-sin^2A

=cos^2A-(1-cos^2A)

= cos^2A-1+cos^2A

= 2cos^2A-1 = RHS.

c)

LHS = sinA(1+tanA)+cosA(1+cotA)

= sinA+sin^2A/cosA + cosA +cos^A/sinA

= sinA +(1-cos^2A)/cosA +cosA +(1-sin^2A)/sinA

= sinA+1/cosA -cosA +cosA +1/sinA - sinA

= 1/sinA +1/cosA

= cosecA+secA = RHS.

=

=

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