If L=(4tan1∘)(4tan2∘)(4tan3∘)…(4tan10∘),M=(tan2∘)(tan4∘)(tan6∘)…(tan20∘), and N=(1+tan1∘)(1−tan1∘)(1+tan2∘)(1−tan2∘)…(1+tan10∘)(1−tan10∘), then the value of L128MN is
Open in App
Solution
Given L=(4tan1∘)(4tan2∘)(4tan3∘)…(4tan10∘)M=(tan2∘)(tan4∘)(tan6∘)…(tan20∘)N=(1+tan1∘)(1−tan1∘)(1+tan2∘)(1−tan2∘)…(1+tan10∘)(1−tan10∘)⇒N=(1−tan21∘)(1−tan22∘)…(1−tan210∘)
We know that tan2∘=2tan1∘1−tan21∘
So, MN=210tan1∘tan2∘tan3∘…tan10∘⇒LMN=410210⇒LMN=210∴L128MN=21027=8