If tan -x-1- tan-x-1- show that 2 -x2-

We have, tan α = x+1 and tan β = x 1 Now, 2 cot (α β) = 2/tan(α β) = 2/tan α tan β/1+ tan α tan β = 2(1+tan α tan β)/tan α tan β = 2[1+(x+1)(x 1)]/x+1 ( ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Compound Angles

If tan α=x+1,...

Question

If tan $α = x + 1, t a n β = x - 1$ . show that 2 cot $(α - β) = x^{2}$ .

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Solution

We have,
$t a n α = x + 1 a n d t a n β = x - 1$
Now, 2 cot $(α - β)$
= $\frac{2}{t a n (α - β)} = \frac{2}{\frac{t a n α - t a n β}{1 + t a n α t a n β}} = \frac{2 (1 + t a n α t a n β)}{t a n α - t a n β} = \frac{2 [1 + (x + 1) (x - 1)]}{x + 1 - (x - 1)} = \frac{2 [1 + x^{2} - 1]}{x + 1 - x + 1} = \frac{2 \times x^{2}}{2} = x^{2}$
$∴ 2 c o t (α - β) = x^{2}$

Hence proved

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If tan α=x+1,tanβ=x−1. show that 2 cot (α−β)=x2.

We have, tanα=x+1andtanβ=x−1 Now, 2 cot (α−β) = 2tan(α−β)=2tanα−tanβ1+tanαtanβ=2(1+tanαtanβ)tanα−tanβ=2[1+(x+1)(x−1)]x+1−(x−1)=2[1+x2−1]x+1−x+1=2×x22=x2 ∴2cot(α−β)=x2 Hence proved

If tan $α = x + 1, t a n β = x - 1$ . show that 2 cot $(α - β) = x^{2}$ .