Question 7tan - - tan90- - - - sec - sec90- - -

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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Standard VI

Biology

Pond Habitat

Question 7tan...

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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Similar questions

Q. Question 7

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

Q. Simplify

\frac{sec (90^{\circ} - θ)}{sin (90^{\circ} - θ)} \times \frac{cos θ}{tan (90^{\circ} - θ)} - sec θ

Q. Evaluate

\frac{cos θ}{sin (90 + θ)} + \frac{sin (- θ)}{sin (180 + θ)} + \frac{tan (90 + θ)}{cot θ}

Prove the following:

(i) $s i n θ s i n (90^{o} - θ) - c o s θ c o s (90^{o} - θ) = 0$
(ii) $\frac{c o s (90^{o} - θ) s e c (90^{o} - θ) t a n θ}{c o s e c (90^{o}) s i n (90^{o} - θ) c o t (90^{o} - θ)} + \frac{t a n (90^{o} - θ)}{c o t θ} = 2$
(iii) $\frac{t a n (90^{o} - A) c o t A}{c o s e c^{2} A} - c o s^{2} A = 0$
(iv) $\frac{c o s (90^{o} - A) s i n (90^{o} - A)}{t a n (90^{o} - A)} = s i n^{2} A$
(v) $s i n (50^{o} + θ) - c o s (40^{o} - θ) + t a n 1^{o} t a n 10^{o} t a n 20^{o} t a n 70^{o} t a n 80^{o} t a n 89^{o} = 1$

Q. The value of

\frac{cos Θ}{sin (90 + Θ)} + \frac{sin (- Θ)}{sin (180 + Θ)} + \frac{tan (90 + Θ)}{cot Θ}

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

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