Question 7 tan-tan90-90-

LHS=tan θ+tan(90^∘ θ) [∵ tan(90^∘ θ )=cot θ ] =tan θ+cot θ=sin θ/cos θ+cos θ/sin θ =sin^2 θ+cos^2 θ/sin θ cos θ [tan θ=sin θ/cos θ and cot θ=cos θ/sin θ ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Ratios of a Right Triangle

Question 7 ta...

Question

Question 7
$t a n θ + t a n (90^{\circ} - θ) = s e c θ . s e c (90^{\circ} - θ)$

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Solution

$L H S = t a n θ + t a n (90^{\circ} - θ) [∵ t a n (90^{\circ} - θ) = c o t θ]$

$= t a n θ + c o t θ = \frac{s i n θ}{c o s θ} + \frac{c o s θ}{s i n θ}$

$= \frac{s i n^{2} θ + c o s^{2} θ}{s i n θ c o s θ} [t a n θ = \frac{s i n θ}{c o s θ} a n d c o t θ = \frac{c o s θ}{s i n θ}]$

$= \frac{1}{s i n θ c o s θ} [∵ s i n^{2} θ + c o s^{2} θ = 1]$

$= s e c θ . c o s e c θ [∵ s i n^{2} θ + c o s^{2} θ = 1]$

$= s e c θ . c o s e c θ [∵ s e c θ = \frac{1}{c o s θ} a n d c o s θ = \frac{1}{s i n θ}]$

$= s e c θ . s e c (90^{\circ} - θ) = R H S [∵ s e c (90^{\circ} - θ) = c o s e c θ]$

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t a n θ + t a n (90^{\circ} - θ) = s e c θ . s e c (90^{\circ} - θ)

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Question 7 tan θ+tan(90∘−θ)=sec θ.sec(90∘−θ)

Question 7
$t a n θ + t a n (90^{\circ} - θ) = s e c θ . s e c (90^{\circ} - θ)$