Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$\left(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\right)=\left(\frac{1-\tan A}{1-\cot A}\right)^{2}=\tan ^{2} A$
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$\left(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\right)=\left(\frac{1-\tan A}{1-\cot A}\right)^{2}=\tan ^{2} A$
$\left(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\right)=\left(\frac{1-\tan A}{1-\cot A}\right)^{2}=\tan ^{2} A$
$\frac{1+\tan ^{2} A}{1+\cot ^{2} A}=\frac{1+\frac{\sin ^{2} A}{\cos ^{2} A}}{1+\frac{\cos ^{2} A}{\sin ^{2} A}}=\frac{\frac{\cos ^{2} A+\sin ^{2} A}{\cos ^{2} A}}{\frac{\sin ^{2} A+\cos ^{2} A}{\sin ^{2} A}}$
$=\frac{1}{\cos ^{2} A}=\frac{\sin ^{2} A}{\sin ^{2} A}$
$=\tan ^{2} A$
$\left(\frac{1-\tan A }{1-\cot A }\right)^{2}=\frac{1+\tan ^{2} A -2 \tan A }{1+\cot ^{2} A -2 \cot A }$
$=\frac{\sec ^{2} A-2 \tan A}{\operatorname{cosec}^{2} A-2 \cot A}$
$=\frac{\frac{1}{\cos ^{2} A}-\frac{2 \sin A}{\cos A}}{\frac{1}{\sin ^{2} A}-\frac{2 \cos A}{\sin A}}=\frac{\frac{1-2 \sin A \cos A}{\cos ^{2} A}}{\frac{1-2 \sin A \cos A}{\sin ^{2} A}}$
$=\frac{\sin ^{2} A }{\cos ^{2} A }=\tan ^{2} A$
Similar Questions
Evaluate:
$\frac{\sin 18^{\circ}}{\cos 72^{\circ}}$
Evaluate:
$\frac{\sin 18^{\circ}}{\cos 72^{\circ}}$
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$(\operatorname{cosec} A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}$
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$(\operatorname{cosec} A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}$
State whether the following are true or false. Justify your answer.
$\sin \theta=\cos \theta$ for all values of $\theta$
State whether the following are true or false. Justify your answer.
$\sin \theta=\cos \theta$ for all values of $\theta$
If $\angle A$ and $\angle B$ are acute angles such that $\cos A =\cos B ,$ then show that $\angle A =\angle B$.
If $\angle A$ and $\angle B$ are acute angles such that $\cos A =\cos B ,$ then show that $\angle A =\angle B$.
If $\cot \theta=\frac{7}{8},$ evaluate:
$(i)$ $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$
$(ii)$ $\cot ^{2} \theta$
If $\cot \theta=\frac{7}{8},$ evaluate:
$(i)$ $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$
$(ii)$ $\cot ^{2} \theta$