Prove that: (secA+cosA)²+(cotA+tanA)²+ (cosecA+sinA)²=9+2tan²A+2cot²A
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Answer:
(secA-cosA)²+(cosecA-sinA)²-(cotA-tanA)²=1
L.H.S.=(secA-cosA)²+(cosecA-sinA)²-(cotA-tanA)²
=sec^2A-2secAcosA+cos^2A+cosec^2A-2cosecAsinA+sin^2A-cot^2A+2cotAtanA-tan^2A
=1+tan^2A-2+1+1-2+2-tan^2A
=1
=R.H.S
Step-by-step explanation:
Answered by
1
Answer:
See the explanation
Step-by-step explanation:
Take LHS and open it
Sec^2A+cos^2A+2+cot^2A+tan^2A+2+cosec^2A+sin^2A+2
2+2+2+1+sec^2A+cot^2A+tan^2A+cosec^2A
7+ (1+tan^2A)+cot^2A+tan^2A+(1+cot^2A)
9+2tan^2A+2cot^2A
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