tan (65^circ - theta) - cot (25^circ + theta) | vedclass

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

Question

(D) We know the trigonometric identities for complementary angles: $	an(90^\circ - A) = \cot A$ and $\sec(90^\circ - A) = \operatorname{cosec} A$.Fir

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) We know the trigonometric identities for complementary angles: $	an(90^\circ - A) = \cot A$ and $\sec(90^\circ - A) = \operatorname{cosec} A$.First,consider the term $	an(65^\circ - 	heta)$:$	an(65^\circ - 	heta) = \cot(90^\circ - (65^\circ - 	heta)) = \cot(25^\circ + 	heta)$.Next,consider the term $\sec(55^\circ - 	heta)$:$\sec(55^\circ - 	heta) = \operatorname{cosec}(90^\circ - (55^\circ - 	heta)) = \operatorname{cosec}(35^\circ + 	heta)$.Now,substitute these into the given expression:$	an(65^\circ - 	heta) - \cot(25^\circ + 	heta) - \sec(55^\circ - 	heta) + \operatorname{cosec}(35^\circ + 	heta)$$= \cot(25^\circ + 	heta) - \cot(25^\circ + 	heta) - \operatorname{cosec}(35^\circ + 	heta) + \operatorname{cosec}(35^\circ + 	heta)$$= 0$.

$\tan (65^\circ - \theta) - \cot (25^\circ + \theta) - \sec (55^\circ - \theta) + \operatorname{cosec}(35^\circ + \theta) = \ldots \ldots \ldots \ldots$ (where,$0 < \theta < 25^\circ$)

Similar Questions

If $\sin \theta + \cos \theta = \sqrt{3}$,then prove that $\tan \theta + \cot \theta = 1$.

Prove that:
If $\tan A = \frac{3}{4},$ then $\sin A \cos A = \frac{12}{25}$

If $\sin \theta + \sin^2 \theta = 1$,then $\cos^2 \theta + \cos^4 \theta = \dots$

If $5 \theta$ is the measure of an acute angle and $\cos \theta = \sin 5 \theta$,then the value of $\theta$ is $\ldots \ldots \ldots \ldots$ (in $^{\circ}$)

The value of $(\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 89^{\circ})$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module