Prove that sin 90- - -1 - sin- - cos-1 - cos 90- - - - 2-

LHS = sin (90^∘ θ)1 + sinθ + cosθ1 cos (90^∘ θ) = cosθ1 + sinθ + cosθ1 sinθ = cosθ(1 sinθ) + cosθ (1 + sinθ)1 sin^2θ = cosθ co ...

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Standard X

Mathematics

Complementary Trigonometric Ratios

Prove that s...

Question

Prove that
$\frac{sin (90^{\circ} - θ)}{1 + sin θ} + \frac{cos θ}{1 - cos (90^{\circ} - θ)} = 2 sec θ$

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Solution

$L H S = \frac{sin (90^{\circ} - θ)}{1 + sin θ} + \frac{cos θ}{1 - cos (90^{\circ} - θ)}$

$= \frac{cos θ}{1 + sin θ} + \frac{cos θ}{1 - sin θ}$

$= \frac{cos θ (1 - sin θ) + cos θ (1 + sin θ)}{1 - {sin}^{2} θ}$

$= \frac{cos θ - cos θ . sin θ + cos θ + cos θ . sin θ}{1 - {sin}^{2} θ}$

$= \frac{2 cos θ}{{cos}^{2} θ}$ $= \frac{2}{cos θ}$ $= 2 sec θ$

$L H S = R H S$

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Similar questions

Prove that:

(i) $s i n θ c o s (90^{\circ} - θ) + s i n (90^{\circ} - θ) c o s θ = 1$

(ii) $\frac{s i n θ}{c o s (90^{\circ} - θ)} + \frac{c o s θ}{s i n (90^{\circ} - θ)} = 2$

(iii) $\frac{s i n θ c o s (90^{\circ} - θ) c o s θ}{s i n (90^{\circ} - θ)} + \frac{c o s θ s i n (90^{\circ} - θ) s i n θ}{c o s (90^{\circ} - θ)} = 1$

(iv) $\frac{c o s (90^{\circ} - θ) s e c (90^{\circ} - θ) t a n θ}{c o s e c (90^{\circ} - θ) s i n (90^{\circ} - θ) c o t (90^{\circ} - θ)} + \frac{t a n (90^{\circ} - θ)}{c o t θ} = 2$

(v) $\frac{c o s (90^{\circ} - θ)}{1 + s i n (90^{\circ} - θ)} + \frac{1 + s i n (90^{\circ} - θ)}{c o s (90^{\circ} - θ)} = 2 c o s e c θ$

(vi) $\frac{s e c (90^{\circ} - θ) c o s e c θ - t a n (90^{\circ} - θ) c o t θ + c o s^{2} 25^{\circ} + c o s^{2} 65^{\circ}}{3 t a n 27^{\circ} t a n 63^{\circ}} = \frac{2}{3}$

(vii) $c o t θ t a n (90^{\circ} - θ) - s e c (90^{\circ} - θ) c o s e c θ + \sqrt{3} t a n 12^{\circ} t a n 60^{\circ} t a n 78^{\circ} = 2$

Q. Evaluate:

\frac{cos θ}{sin (90^{\circ} - θ)} + \frac{sin θ}{cos (90^{\circ} - θ)}

Q. Prove that

\frac{sin θ - cos θ + 1}{sin θ + cos θ - 1} = \frac{1}{sec θ - tan θ}

, using the identity

{sec}^{2} θ = 1 + {tan}^{2} θ

The sum of the squares of sin $θ$ + sin ( $90^{\circ} - θ$ ) and cos $θ$ - cos ( $90^{\circ} - θ$ ) is equal to 2.

$\frac{1 + s i n (90^{\circ} - θ) - {c o s}^{2} (90^{\circ} - θ)}{c o s (90^{\circ} - θ) [1 + s i n (90^{\circ} - θ)]} =$

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Prove thatsin(90∘−θ)1+sinθ+cosθ1−cos(90∘−θ)=2secθ

LHS=sin(90∘−θ)1+sinθ+cosθ1−cos(90∘−θ)=cosθ1+sinθ+cosθ1−sinθ=cosθ(1−sinθ)+cosθ(1+sinθ)1−sin2θ=cosθ−cosθ.sinθ+cosθ+cosθ.sinθ1−sin2θ=2cosθcos2θ =2cosθ =2secθLHS=RHS

Prove that
$\frac{sin (90^{\circ} - θ)}{1 + sin θ} + \frac{cos θ}{1 - cos (90^{\circ} - θ)} = 2 sec θ$