Prove that - - sin- -cos- - -2

(secθ #160; + cos ecθ )(sinθ #160; + cosθ ) =(1cosθ+1sin θ)(sin θ+cos θ) =(sin θ+cos θ)(sin θ+cos θ)sin θ.cos θ =sin^2 θ + cos^2 θ +2sin ...

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Standard X

Mathematics

Trigonometric Ratios of a Right Triangle

Prove that ...

Question

Prove that $(sec θ + csc θ) (sin θ + cos θ) = sec θ csc θ + 2$

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Solution

$(s e c θ + cos e c θ) (s i n θ + c o s θ)$

$= (\frac{1}{c o s θ} + \frac{1}{s i n θ}) (s i n θ + c o s θ)$

$= \frac{(s i n θ + c o s θ) (s i n θ + c o s θ)}{s i n θ . c o s θ}$

$= \frac{s i n^{2} θ + c o s^{2} θ + 2 s i n θ c o s θ}{s i n θ c o s θ}$

$= \frac{1 + 2 s i n θ c o s θ}{s i n θ c o s θ}$

$= \frac{1}{s i n θ c o s θ} + \frac{2 s i n θ c o s θ}{s i n θ c o s θ}$

$= s e c θ c o s e c θ + 2$

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Similar questions

Q. Prove that :

\frac{csc θ}{1 + sec θ} + \frac{1 + sec θ}{csc θ} = \frac{sec θ csc θ + sin θ cos θ + 2 sin θ}{1 + cos θ}

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

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

Q. If

{tan}^{2} θ = (1 - e^{2})

, prove that

sec θ + {tan}^{3} θ csc θ = {(2 - e^{2})}^{3 / 2}

Q. Prove the following trigonometric identities.

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

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Prove that (secθ+cscθ)(sinθ+cosθ)=secθcscθ+2

(secθ+cosecθ)(sinθ+cosθ)=(1cosθ+1sinθ)(sinθ+cosθ)=(sinθ+cosθ)(sinθ+cosθ)sinθ.cosθ=sin2θ+cos2θ+2sinθcosθsinθcosθ=1+2sinθcosθsinθcosθ=1sinθcosθ+2sinθcosθsinθcosθ=secθcosecθ+2

Prove that $(sec θ + csc θ) (sin θ + cos θ) = sec θ csc θ + 2$