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Basic Trigonometric Identities

1+tanαtanβ2+t...

Question

$(1 + t a n α t a n β)^{2} + (t a n α - t a n β)^{2} = s e c^{2} α s e c^{2} β$

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Solution

$L H S = (1 + t a n α t a n β)^{2} + (t a n α - t a n β)^{2}$

$= 1 + (t a n α + t a n β)^{2} + 2.1 t a n α t a n β + (t a n α)^{2} + (t a n β)^{2} + 2 t a n α . t a n β$

[Using $(a + b)^{2} = a^{2} + b^{2} + 2 a b a n d (a - b)^{2} = a^{2} + b^{2} - 2 a b]$

$= 1 + t a n^{2} α - t a n^{2} β + 2 t a n α t a n β + t a n^{2} α + t a n^{2} β - 2 t a n α t a n β$

$= 1 + t a n^{2} α + t a n^{2} α - t a n^{2} β + t a n^{2} β$

$= s e c^{2} α + t a n^{2} β (1 + t a n^{2} α) [∵ 1 + t a n^{2} α = s e c^{2} α]$

$= s e c^{2} α + t a n^{2} β . s e c^{2} α$

$= s e c^{2} α (1 + t a n^{2} β)$

$= s e c^{2} α . s e c^{2} β [∵ 1 + t a n^{2} β = s e c^{2} β]$

=RHS Hence proved.

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Similar questions

Q. Prove the following identities (1-17)

{(1 + \tan α \tan β)}^{2} + {(\tan α - \tan β)}^{2} = \sec^{2} α \sec^{2} β

(1 + t a n α t a n β)^{2} + (t a n α - t a n β)^{2}

{(1 + tan α tan β)}^{2} + {(tan α - tan β)}^{2} = {sec}^{m} α {sec}^{m} β

m =

[\begin{matrix} 1 & - tan α / 2 tan α / 2 & 1 \end{matrix}] [\begin{matrix} tan α / 2 & 1 1 & 0 \end{matrix}]

sec α = \frac{5}{4}

\frac{1 - tan α}{1 + tan α}

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