please answer these both asap
Answers
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.
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.