Question

please answer these both asap ​

Answer 1

Answer:

Proved below.

Step-by-step explanation:

From the properties of trigonometry :

• tanA = sinA / cosA
• sin^2 A + cos^2 A = 1
• sinA = 1 / cosecA
• cosA = 1 / secA
• tanA = 1 / cotA

Question 1 : sec^2 A + cosec^2 A

⇒ sec^2 A + cosec^2 A

⇒ ( secA )^2 + ( cosecA )^2

⇒ ( 1 / cosA )^2 + ( 1 / sinA )^2

⇒ 1 / cos^2 A + 1 / sin^2 A

⇒ ( sin^2 A + cos^2 A ) / ( sin^2 A.cos^2 A )

⇒ 1 / ( sin^2 A cos^2 A )              { sin^2 A + cos^2 A = 1 }

⇒ ( 1 / sin^2 A )( 1 / cos^2 A )

⇒ cosec^2 A . sec^2 A

⇒ sec^2A.cosec^2A.

        Proved.

   

         Question 2 : tan^2 A + cot^2 + 2

⇒ tan^2 + cot^2 A + 2( 1 )

⇒ tan^2 A + cot^2 A + 2( tanA x cotA )        { tanA.cotA = 1 }

⇒ ( tanA + cotA )^2                    { Using a^2 + b^2 + 2ab = ( a + b )^2 }

⇒ { ( sinA / cosA ) + ( cosA / sinA ) }^2

⇒ [ ( sin^2 A + cos^2 A ) / sinAcosA ]^2

⇒ { 1 / sinAcosA }^2

⇒ ( secAcosecA )^2

⇒ sec^2 A . cosec^2 A

      Proved.

Answer 2

Step-by-step explanation:

Hi Devinder pal

Here's your answer.

1. To prove

sec^2A+cosec^2A=sec^2Acosec^2A

sec^2A+cosec^2A = LHS

secA=1/cos A and cosecA=1/sinA

1/cos^2A+1/sin^2A

taking LCM

(sin^2A+cos^2A)/cos^2Asin^2A

As sin^2A+cos^2A=1

1/cos^2Asin^2A

(1/cos^2A)×(1/sin^2A)

sec^2Acosec^2A

RHS=sec^2Acosec^2A

LHS=RHS

Hence proved.

2. To prove

tan^2A+cot^2A+2=sec^2Acosec^2A

tan^2A+cot^2A+2=LHS

As tanA=sinA/cosA and cotA=cosA/sinA

sin^2A/cos^2A+cos^2A/sin^2A+2

taking LCM

(sin^4A+cos^4A+2cos^2Asin^2A)/sin^2Acos^2A

sin^4A+cos^4A+2cos^2Asin^2A can be written as

(sin^2A+cos^2A)^2

(sin^2A+cos^2A)^2/sin^2cos^2

As sin^2A+cos^2A=1

1/cos^2Asin^2A

(1/cos^2A)×(1/sin^2A)

sec^2Acosec^2A

RHS=sec^2Acosec^2A

LHS=RHS

Hence proved.

Hope, it helps.