Prove following identities, where angles involved are acute angles for which expressions are defined

English

Gujarati

Hindi

8. Introduction to Trigonometry

hard

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

$\left(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\right)=\left(\frac{1-\tan A}{1-\cot A}\right)^{2}=\tan ^{2} A$

Option A

Option B

Option C

Option D

Solution

$\left(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\right)=\left(\frac{1-\tan A}{1-\cot A}\right)^{2}=\tan ^{2} A$

$\frac{1+\tan ^{2} A}{1+\cot ^{2} A}=\frac{1+\frac{\sin ^{2} A}{\cos ^{2} A}}{1+\frac{\cos ^{2} A}{\sin ^{2} A}}=\frac{\frac{\cos ^{2} A+\sin ^{2} A}{\cos ^{2} A}}{\frac{\sin ^{2} A+\cos ^{2} A}{\sin ^{2} A}}$

$=\frac{1}{\cos ^{2} A}=\frac{\sin ^{2} A}{\sin ^{2} A}$

$=\tan ^{2} A$

$\left(\frac{1-\tan A }{1-\cot A }\right)^{2}=\frac{1+\tan ^{2} A -2 \tan A }{1+\cot ^{2} A -2 \cot A }$

$=\frac{\sec ^{2} A-2 \tan A}{\operatorname{cosec}^{2} A-2 \cot A}$

$=\frac{\frac{1}{\cos ^{2} A}-\frac{2 \sin A}{\cos A}}{\frac{1}{\sin ^{2} A}-\frac{2 \cos A}{\sin A}}=\frac{\frac{1-2 \sin A \cos A}{\cos ^{2} A}}{\frac{1-2 \sin A \cos A}{\sin ^{2} A}}$

$=\frac{\sin ^{2} A }{\cos ^{2} A }=\tan ^{2} A$

Standard 10

Mathematics

If $\sec 4 A =\operatorname{cosec}\left( A -20^{\circ}\right),$ where $4 A$ is an acute angle, find the value of $A$. (in $^{\circ}$)

easy

State whether the following are true or false. Justify your answer.

The value of $\cos \theta$ increases as $\theta$ increases

easy

Evaluate the following:

$2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ}$

easy

State whether the following are true or false. Justify your answer.

$\cot$ $A$ is not defined for $A =0^{\circ}$

easy

If $A , B$ and $C$ are interior angles of a triangle $ABC ,$ then show that

$\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}$

medium

Prove the following identities, where the angles involved are acute angles for which the expressions are defined. $\left(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\right)=\left(\frac{1-\tan A}{1-\cot A}\right)^{2}=\tan ^{2} A$

Solution

Similar Questions

If $\sec 4 A =\operatorname{cosec}\left( A -20^{\circ}\right),$ where $4 A$ is an acute angle, find the value of $A$. (in $^{\circ}$)

State whether the following are true or false. Justify your answer. The value of $\cos \theta$ increases as $\theta$ increases

Evaluate the following: $2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ}$

State whether the following are true or false. Justify your answer. $\cot$ $A$ is not defined for $A =0^{\circ}$

If $A , B$ and $C$ are interior angles of a triangle $ABC ,$ then show that $\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}$

Start a Free Trial Now

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

$\left(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\right)=\left(\frac{1-\tan A}{1-\cot A}\right)^{2}=\tan ^{2} A$

State whether the following are true or false. Justify your answer.

The value of $\cos \theta$ increases as $\theta$ increases

Evaluate the following:

$2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ}$

State whether the following are true or false. Justify your answer.

$\cot$ $A$ is not defined for $A =0^{\circ}$

If $A , B$ and $C$ are interior angles of a triangle $ABC ,$ then show that

$\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}$