Question

Prove that: (secA+cosA)²+(cotA+tanA)²+ (cosecA+sinA)²=9+2tan²A+2cot²A

Answer 1

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:

Answer 2

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