prove that:(1+sec2A)(1+sec4A)(1+sec8A)=tan8A*cotA write full solution not just hint please
chandrika yadav , 6 Years ago
Grade 10
Trigonometry> prove that: (1+sec2A)(1+sec4A)(1+sec8A)=t...
1 Answers
Arun
Last Activity: 6 Years ago
recall your half-angle formula:
tan(x/2) = sinx/(1+cosx)
Now, we have
1+sec2A = (1+cos2A)/cos2A
= 1/[cos2A/(1+cos2A)]
= 1/[sin2A/(1+cos2A) * cos2A/sin2A]
= 1/(tanA cot2A)
= tan2A/tanA
Now e can see that
1+sec4A = tan4A/tan2A
1+sec8A = tan8A/tan4A
tan(x/2) = sinx/(1+cosx)
Now, we have
1+sec2A = (1+cos2A)/cos2A
= 1/[cos2A/(1+cos2A)]
= 1/[sin2A/(1+cos2A) * cos2A/sin2A]
= 1/(tanA cot2A)
= tan2A/tanA
Now e can see that
1+sec4A = tan4A/tan2A
1+sec8A = tan8A/tan4A
Now multiply and get the result
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