Evaluate the following :
Question:

Evaluate the following :

(i) $\left(\frac{\sin 49^{\circ}}{\cos 41^{\circ}}\right)^{2}+\left(\frac{\cos 41^{\circ}}{\sin 49^{\circ}}\right)^{2}$

(ii) $\cos 48^{\circ}-\sin 42^{\circ}$

(iii) $\frac{\cot 40^{\circ}}{\tan 50^{\circ}}-\frac{1}{2}\left(\frac{\cos 35^{\circ}}{\sin 55^{\circ}}\right)$

(iv) $\left(\frac{\sin 27^{*}}{\cos 63^{*}}\right)^{2}-\left(\frac{\cos 63^{*}}{\sin 27^{*}}\right)^{2}$

(v) $\frac{\tan 35^{\circ}}{\cot 55^{\circ}}+\frac{\cot 78^{\circ}}{\tan 12^{\circ}}-1$

(vi) $\frac{\sec 70^{\circ}}{\operatorname{cosec} 20^{\circ}}+\frac{\sin 59^{\circ}}{\cos 31^{\circ}}$

(vii) $\operatorname{cosec} 31^{\circ}-\sec 59^{\circ}$

(viii) $\left(\sin 72^{\circ}+\cos 18^{\circ}\right)\left(\sin 72^{\circ}-\cos 18^{\circ}\right)$

(ix) $\sin 35^{\circ} \sin 55^{\circ}-\cos 35^{\circ} \cos 55^{\circ}$

(x) $\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}$

 

(xi) $\sec 50^{\circ} \sin 40^{\circ}+\cos 40^{\circ} \operatorname{cosec} 50^{\circ}$

Solution:

(i) We have to find: $\left(\frac{\sin 49^{\circ}}{\cos 41^{\circ}}\right)^{2}+\left(\frac{\cos 41^{\circ}}{\sin 49^{\circ}}\right)^{2}$

Since $\frac{\sec 70^{\circ}}{\operatorname{cosec} 20^{\circ}}+\frac{\sin 59^{\circ}}{\cos 31^{\circ}} \sin \left(90^{\circ}-\theta\right)=\cos \theta$ and $\cos \left(90^{\circ}-\theta\right)=\sin \theta$

So

$\left(\frac{\sin \left(90^{\circ}-41^{\circ}\right)}{\cos 41^{\circ}}\right)^{2}+\left(\frac{\cos \left(90^{\circ}-49^{\circ}\right)}{\sin 49^{\circ}}\right)^{2}=\left(\frac{\cos 41^{\prime}}{\cos 41^{\circ}}\right)^{2}+\left(\frac{\sin 49^{\circ}}{\sin 49^{\circ}}\right)^{2}$

$=1+1$

 

$=2$

So value of $\left(\frac{\sin 49^{\prime \prime}}{\cos 41^{\circ}}\right)^{2}+\left(\frac{\cos 41^{\circ}}{\sin 49^{\circ}}\right)^{2}$ is 2

(ii) We have to find: $\cos 48^{\circ}-\sin 42^{\circ}$

Since $\cos \left(90^{\circ}-\theta\right)=\sin \theta$.So

$\cos 48^{\circ}-\sin 42^{\circ}=\cos \left(90^{\circ}-42^{\circ}\right)-\sin 42^{\circ}$

$=\sin 42^{\circ}-\sin 42^{\circ}$

 

$=0$

So value of $\cos 48^{\circ}-\sin 42^{\circ}$ is 0

(iii) We have to find:

$\frac{\cot 40^{\circ}}{\tan 50^{\circ}}-\frac{1}{2}\left(\frac{\cos 35^{\circ}}{\sin 55^{\circ}}\right)$

Since $\cot \left(90^{\circ}-\theta\right)=\tan \theta$ and $\cos \left(90^{\circ}-\theta\right)=\sin \theta$

$\frac{\cot 40^{\circ}}{\tan 50^{\circ}}-\frac{1}{2}\left(\frac{\cos 35^{\circ}}{\sin 55^{\circ}}\right)=\frac{\cot \left(90^{\circ}-50^{\circ}\right)}{\tan 50^{\circ}}-\frac{1}{2}\left(\frac{\cos \left(90^{\circ}-55^{\circ}\right)}{\sin 55^{\circ}}\right)$

$=\frac{\tan 50^{\circ}}{\tan 50^{\circ}}-\frac{1}{2}\left(\frac{\sin 55^{\circ}}{\sin 55^{\circ}}\right)$

$=1-\frac{1}{2}$

$=\frac{1}{2}$

So value of $\frac{\cot 40^{\circ}}{\tan 50^{\circ}}-\frac{1}{2}\left(\frac{\cos 35^{\circ}}{\sin 55^{\circ}}\right)$ is $\frac{1}{2}$

(iv) We have to find: $\left(\frac{\sin 27^{\circ}}{\cos 63^{\circ}}\right)^{2}-\left(\frac{\cos 63^{\circ}}{\sin 27^{\circ}}\right)^{2}$

Since $\sin \left(90^{\circ}-\theta\right)=\cos \theta$ and $\cos \left(90^{\circ}-\theta\right)=\sin \theta$

$\left(\frac{\sin 27}{\cos 63^{\circ}}\right)^{2}-\left(\frac{\cos 63}{\sin 27}\right)^{2}=\left(\frac{\sin \left(90^{\circ}-63^{\prime}\right)}{\cos 63^{\circ}}\right)^{2}-\left(\frac{\cos \left(90^{\circ}-27^{\circ}\right)}{\sin 27}\right)^{2}$

$=\left(\frac{\cos 63^{\circ}}{\cos 63^{\circ}}\right)^{2}-\left(\frac{\sin 27}{\sin 27}\right)^{2}$

$=1-1$

 

$=0$

So value of $\left(\frac{\sin 27}{\cos 63^{\circ}}\right)^{2}-\left(\frac{\cos 63^{\circ}}{\sin 27}\right)^{2}$ is 0

(v) We have to find:

$\frac{\tan 35^{\circ}}{\cot 55^{\circ}}+\frac{\cot 78^{\circ}}{\tan 12^{\circ}}-1$

Since $\tan \left(90^{\circ}-\theta\right)=\cot \theta$ and $\cot \left(90^{\circ}-\theta\right)=\tan \theta$

$=1$

So value of $\frac{\tan 35^{\circ}}{\cot 55^{\circ}}+\frac{\cot 78^{\circ}}{\tan 12^{\circ}}$ is $\square$

(vi) We have to find: $\frac{\sec 70^{\circ}}{\operatorname{cosec} 20^{\circ}}+\frac{\sin 59^{\circ}}{\cos 31^{\circ}}$

Since $\frac{\sec 70^{\circ}}{\operatorname{cosec} 20^{\circ}}+\frac{\sin 59^{\circ}}{\cos 31^{\circ}}$ and $\sec \left(90^{\circ}-\theta\right)=\operatorname{cosec} \theta$

So

$\frac{\sec 70^{\circ}}{\operatorname{cosec} 20^{\circ}}+\frac{\sin 59^{\circ}}{\cos 31^{\prime}}=\frac{\sec \left(90^{\circ}-20^{\circ}\right)}{\operatorname{cosec} 20^{\circ}}+\frac{\sin \left(90^{\circ}-31^{\circ}\right)}{\cos 31^{\prime}}$

$=\frac{\operatorname{cosec} 20^{\circ}}{\operatorname{cosec} 20^{\circ}}+\frac{\cos 31^{\circ}}{\cos 31^{\circ}}$

$=1+1$

 

$=2$

So value of $\frac{\sec 70^{\circ}}{\operatorname{cosec} 20^{\circ}}+\frac{\sin 59^{\circ}}{\cos 31^{\circ}}$ is 2

(vii) We have to find: $\operatorname{cosec} 31^{\circ}-\sec 59^{\circ}$

Since $\operatorname{cosec}\left(90^{\circ}-\theta\right)=\sec \theta .$ So

$=\operatorname{cosec} 31^{\circ}-\sec 59^{\circ}$

$=\operatorname{cosec}\left(90^{\circ}-59^{\circ}\right)-\sec 59^{\circ}$

$=\sec ^{\circ} 59^{\circ}-\sec 59^{\circ}$

$=0$

So value of $\operatorname{cosec} 31^{\circ}-\sec 59^{\circ}$ is 0

(viii) We have to find: $\left(\sin 72^{\circ}+\cos 18^{\circ}\right)\left(\sin 72^{\circ}-\cos 18^{\circ}\right)$

Since $\sin \left(90^{\circ}-\theta\right)=\cos \theta$. So

$\left(\sin 72^{\circ}+\cos 18^{\circ}\right)\left(\sin 72^{\circ}-\cos 18^{\circ}\right)=\left(\sin 72^{\circ}\right)^{2}-\left(\cos 18^{\circ}\right)^{2}$

$=\left[\sin \left(90^{\circ}-18^{\prime}\right)\right]^{2}-\left(\cos 18^{\circ}\right)^{2}$

$=\left(\cos 18^{\circ}\right)^{2}-\left(\cos 18^{\circ}\right)^{2}$

$=\cos ^{2} 18-\cos ^{2} 18$

$=0$

So value of $\left(\sin 72^{\circ}+\cos 18^{\circ}\right)\left(\sin 72^{\circ}-\cos 18^{\circ}\right)$ is 0

(ix) We find: $\sin 35^{\circ} \sin 55^{\circ}-\cos 35^{\circ} \cos 55^{\circ}$

Since $\sin \left(90^{\circ}-\theta\right)=\cos \theta$ and $\cos \left(90^{\circ}-\theta\right)=\sin \theta$

$\sin 35^{\circ} \sin 55^{\circ}-\cos 35^{\circ} \cos 55^{\circ}=\sin \left(90^{\circ}-55^{\circ}\right) \sin 55^{\circ}-\cos \left(90^{\circ}-55^{\circ}\right) \cos 55^{\circ}$

$=\cos 55^{\circ} \sin 55^{\circ}-\sin 55^{\circ} \cos 55^{\circ}$

$=1-1$

 

$=0$

So value of $\sin 35^{\circ} \sin 55^{\circ}-\cos 35^{\circ} \cos 55^{\circ}$ is 0

(x) We have to find $\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}$

Since $\tan \left(90^{\circ}-\theta\right)=\cot \theta$. So

$\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}=\tan \left(90^{\circ}-42^{\circ}\right) \tan \left(90^{\circ}-67^{\circ}\right) \tan 42^{\circ} \tan 67^{\circ}-$

$=\cot 42^{\circ} \cot 67^{\circ} \tan 42^{\circ} \tan 67^{\circ}$

$=\left(\tan 67^{\circ} \cot 67^{\circ}\right)\left(\tan 42^{\prime} \cot 42^{\circ}\right)$

$=1 \times 1$

 

$=1$

So value of $\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}$ is

(xi) We find to find $\sec 50^{\circ} \sin 40^{\circ}+\cos 40^{\circ} \operatorname{cosec} 50^{\circ}$

Since $\cos \left(90^{\circ}-\theta\right)=\sin \theta, \sec \left(90^{\circ}-\theta\right)=\operatorname{cosec} \theta$ and $\sin \theta \cdot \operatorname{cosec} \theta=1$. So

$\sec 50^{\circ} \sin 40^{\circ}+\cos 40^{\circ} \operatorname{cosec} 50^{\circ}=\sec \left(90^{\circ}-40^{\circ}\right) \sin 40^{\circ}+\cos \left(90^{\circ}-50^{\circ}\right) \operatorname{cosec} 50^{\circ}$

$=\operatorname{cosec} 40^{\circ} \sin 40^{\circ}+\sin 50^{\circ} \operatorname{cosec} 50^{\circ}$

$=1+1$

 

$=2$

So value of $\sec 50^{-} \sin 40^{\circ}+\cos 40^{\circ} \operatorname{cosec} 50^{-}$is 2

 

 

Administrator

Leave a comment

Please enter comment.
Please enter your name.