Sin-cos-sin90- cosec-cos90- -

The correct option is B 1 sinθ cosθ [sin(90^∘ θ) cosecθ + cos(90^∘ θ) sec θ] = sinθ cosθ [cosθ cosecθ + sinθ sec θ] ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Trigonometric Identities

sinθcosθ[sin9...

Question

$s i n θ$ $c o s θ$ [ $s i n (90^{\circ} - θ)$ $c o s e c θ$ + $c o s (90^{\circ} - θ)$ $s e c θ$ ]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

-1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
1

$s i n θ$ $c o s θ$ [ $s i n (90^{\circ} - θ)$ $c o s e c θ$ + $c o s (90^{\circ} - θ)$ $s e c θ$ ]

= $s i n θ$ $c o s θ$ [ $c o s θ$ $c o s e c θ$ + $s i n θ$ $s e c θ$ ]

= $s i n θ$ $c o s θ$ [ $\frac{c o s θ}{s i n θ}$ + $\frac{s i n θ}{c o s θ}$ ]

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

= 1

Suggest Corrections

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. Prove that

\frac{cos 9^{0} + sin 9^{0}}{cos 9^{0} - sin 9^{0}} = cot 36^{0}

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

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

Q. Evaluate:

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

Join BYJU'S Learning Program

sinθ cosθ [sin(90∘−θ) cosecθ + cos(90∘−θ) secθ]

The correct option is C 1 sinθ cosθ [sin(90∘−θ) cosecθ + cos(90∘−θ) secθ] = sinθ cosθ [cosθ cosecθ + sinθ secθ] = sinθ cosθ [cosθsinθ + sinθcosθ] = sinθ cosθ [(sin2θ+cos2θ)sinθcosθ] = 1

$s i n θ$ $c o s θ$ [ $s i n (90^{\circ} - θ)$ $c o s e c θ$ + $c o s (90^{\circ} - θ)$ $s e c θ$ ]