Given,
cotα=1,secβ=3−5
So, cosβ=5−3,tanβ=3−4 and tanα=1
Now using formula tan(α+β)=1−tanαtanβtanα+tanβ
⇒tan(α+β)=1+34×11−34=7−1
We know that tanθ is negative in {2}^{nd}&{4}^{th} quadrant but as π<α<23π, so α+β will lie in IVth quadrant.
If cotα=1 and secβ=−35, where π<α<23π and 2π<β<π, then the value of tan(α+β) and the quadrant in which α+β lies, respectively are
Held on 28 Jun 2022 · Verified 6 Jul 2026.
−71 and IVth quadrant
7 and Ist quadrant
−7 and IVth quadrant
71 and Ist quadrant
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