sec 55^{circ} cdot sin 35^{circ} + cos 35^{ci | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We know that $\sec 	heta = \frac{1}{\cos 	heta}$ and $\operatorname{cosec} 	heta = \frac{1}{\sin 	heta}$.Also,$\sin(90^{\circ} - 	heta) = \co

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) We know that $\sec 	heta = \frac{1}{\cos 	heta}$ and $\operatorname{cosec} 	heta = \frac{1}{\sin 	heta}$.Also,$\sin(90^{\circ} - 	heta) = \cos 	heta$ and $\cos(90^{\circ} - 	heta) = \sin 	heta$.Given expression: $\sec 55^{\circ} \cdot \sin 35^{\circ} + \cos 35^{\circ} \cdot \operatorname{cosec} 55^{\circ}$.Convert $55^{\circ}$ to $(90^{\circ} - 35^{\circ})$:$= \sec(90^{\circ} - 35^{\circ}) \cdot \sin 35^{\circ} + \cos 35^{\circ} \cdot \operatorname{cosec}(90^{\circ} - 35^{\circ})$$= \operatorname{cosec} 35^{\circ} \cdot \sin 35^{\circ} + \cos 35^{\circ} \cdot \sec 35^{\circ}$Since $\operatorname{cosec} 35^{\circ} = \frac{1}{\sin 35^{\circ}}$ and $\sec 35^{\circ} = \frac{1}{\cos 35^{\circ}}$:$= \left(\frac{1}{\sin 35^{\circ}}ight) \cdot \sin 35^{\circ} + \cos 35^{\circ} \cdot \left(\frac{1}{\cos 35^{\circ}}ight)$$= 1 + 1 = 2$.

$\sec 55^{\circ} \cdot \sin 35^{\circ} + \cos 35^{\circ} \cdot \operatorname{cosec} 55^{\circ} = \ldots \ldots \ldots \ldots$

Similar Questions

If $\sin x = \sin 60^{\circ} \cdot \cos 30^{\circ} - \cos 60^{\circ} \cdot \sin 30^{\circ}$,then $x = \ldots$ (in $^{\circ}$)

In $\Delta ABC$,$m \angle C = 90^{\circ}$ and $\cos B = \frac{1}{2}$,then $\operatorname{cosec} A = \ldots$

If $\sec \theta = \frac{13}{5}$,then $\cos \theta = \ldots \ldots \ldots \ldots$

Show that $\frac{\cos ^{2}\left(45^{\circ}+\theta\right)+\cos ^{2}\left(45^{\circ}-\theta\right)}{\tan \left(60^{\circ}+\theta\right) \tan \left(30^{\circ}-\theta\right)}=1$

Prove that,$\tan \theta + \tan (90^{\circ} - \theta) = \sec \theta \sec (90^{\circ} - \theta)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module