Given:The value of the expression \( \left(\cos ^{2} 23^{\circ}-\sin ^{2} 67^{\circ}\right) \) is positive.To do:We have to find whether the given statement is true or false.Solution:$\cos ^{2} 23^{\circ}-\sin ^{2} 67^{\circ}=(\cos 23^{\circ}-\sin 67^{\circ})(\cos 23^{\circ}+\sin 67^{\circ})$          [Since $(a^{2}-b^{2})=(a-b)(a+b)$]$=[\cos 23^{\circ}-\sin(90^{\circ}-23^{\circ})](\cos 23^{\circ}+\sin 67^{\circ})$$=(\cos 23^{\circ}-\cos 23^{\circ})(\cos 23^{\circ}+\sin 67^{\circ})$           [Since $\sin (90^{\circ}-\theta)=\cos \theta]$$=0 \times (\cos 23^{\circ}+\sin 67^{\circ})$$=0$which is neither positive nor negative.The given statement is false.

Given:\( \frac{\tan 47^{\circ}}{\cot 43^{\circ}}=1 \)To do:We have to find whether the given statement is true or false.Solution:We know that,$\tan (90^o-\theta)=\cot \theta$Therefore,$\frac{\tan 47^{\circ}}{\cot 43^{\circ}}=\frac{\tan(90^{\circ}-43^{\circ})}{\cot 43^{\circ}}$$=\frac{\cot 43^{\circ}}{\cot 43^{\circ}}$$=1$The given statement is true.

Given: A pole $6\ m$ high casts a shadow $2\sqrt{3}\ m$ long on the ground.To do: We have to find the sun’s elevation.Solution:Let height $=6\ m$length of shadow $=2\sqrt{3}\ m$$\theta$ is angle of elevation$tan\theta=\frac{height}{shadow-length}$$=\frac{6}{2\sqrt{3}}$$=\sqrt{3}$$=tan60^o$$\therefore \theta=60^o$Thus, angle of elevation is $60^o$

Given:\( \sin \left(45^{\circ}+\theta\right)-\cos \left(45^{\circ}-\theta\right) \)To do:We have to find the value of \( \sin \left(45^{\circ}+\theta\right)-\cos \left(45^{\circ}-\theta\right) \).Solution:We know that,$\cos(90^o - \theta) = \sin \theta$Therefore,$\sin (45^{\circ}+\theta)-\cos (45^{\circ}-\theta)= \cos[90^o- (45^o + \theta)] - \cos(45^o- \theta)$$= \cos (45^o - \theta) - \cos (45^o - \theta)$$= 0$

Given:\( \sin \theta-\cos \theta=0 \) To do:We have to find the value of \( \left(\sin ^{4} \theta+\cos ^{4} \theta\right) \).Solution:We know that,$\tan \theta=\frac{\sin \theta}{\cos \theta}$$\tan 45^{\circ}=1$Therefore,$\sin \theta-\cos \theta=0$$\sin \theta=\cos \theta$$\frac{\sin \theta}{\cos \theta}=1$$\tan \theta=1$$\tan \theta=\tan 45^{\circ}$$\Rightarrow \theta=45^{\circ}$This implies,$\sin ^{4} \theta+\cos ^{4} \theta=\sin ^{4} 45^{\circ}+\cos ^{4} 45^{\circ}$$=(\frac{1}{\sqrt{2}})^{4}+(\frac{1}{\sqrt{2}})^{4}$         (Since $\sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}$)$=\frac{1}{4}+\frac{1}{4}$$=\frac{2}{4}$$=\frac{1}{2}$

Given:\( 4 \tan \theta=3 \)To do:We have to find the value of \( \left(\frac{4 \sin \theta-\cos \theta}{4 \sin \theta+\cos \theta}\right) \).Solution:$4 \tan \theta=3$$\tan \theta=\frac{3}{4}$Therefore,$\frac{4 \sin \theta-\cos \theta}{4 \sin \theta+\cos \theta}$Dividing both numerator and denominator by $\cos \theta$, we get,$\frac{4 \sin \theta-\cos \theta}{4 \sin \theta+\cos \theta}=\frac{4 \frac{\sin \theta}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}+1}$$=\frac{4 \tan \theta-1}{4 \tan \theta+1}$          [Since $\tan \theta=\frac{\sin \theta}{\cos \theta}$]$=\frac{4(\frac{3}{4})-1}{4(\frac{3}{4})+1}$           ($\tan \theta=\frac{3}{4}$)$=\frac{3-1}{3+1}$$=\frac{2}{4}$$=\frac{1}{2}$

Given:\( \left[\frac{\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}}{\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}\right] \)To do:We have to find the value of the expression \( \left[\frac{\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}}{\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}\right] \).Solution:We know that, $\sin \left(90^{\circ}-\theta\right)=\cos \theta$$\cos \left(90^{\circ}-\theta\right)=\sin \theta$$\sin ^{2} \theta+\cos ^{2} \theta=1$Therefore, $[\frac{\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}}{\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}]=\frac{\sin ^{2} 22^{\circ}+\sin ^{2}(90^{\circ}-22^{\circ})}{\cos ^{2}(90^{\circ}-68^{\circ})+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin (90^{\circ}-63^{\circ})$$=\frac{\sin ^{2} 22^{\circ}+\cos ^{2} 22^{\circ}}{\sin ^{2} 68^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} . \cos 63^{\circ}$$=\frac{1}{1}+(\sin ^{2} 63^{\circ}+\cos ^{2} 63^{\circ})$$=1+1$$=2$Read More

Given:\( \sin \alpha=\frac{1}{2} \) and \( \cos \beta=\frac{1}{2} \)To do:We have to find the value of \( (\alpha+\beta) \).Solution:  $\sin \alpha =\frac{1}{2}$$=\sin 30^{\circ}$          [Since \sin 30^{\circ}=\frac{1}{2}$]This implies,$\alpha=30^{\circ}$$\cos \beta =\frac{1}{2}$$=\cos 60^{\circ}$        [Since \cos 60^{\circ}=\frac{1}{2}$]This implies,$\beta=60^{\circ}$Therefore,$\alpha+\beta=30^{\circ}+60^{\circ}$$=90^{\circ}$The value of \( (\alpha+\beta) \) is $90^{\circ}$.

Given:\( \sin \mathrm{A}+\sin ^{2} \mathrm{~A}=1 \)To do:We have to find the value of the expression \( \left(\cos ^{2} \mathrm{~A}+\cos ^{4} \mathrm{~A}\right) \).Solution:  We know that,$\sin ^{2} A+\cos ^{2} A=1$Therefore,$\sin A+\sin ^{2} A =1$$\sin A =1-\sin ^{2} A$$=\cos ^{2} A$Squaring on both sides, we get,$\sin ^{2} A=\cos ^{4} A$$1-\cos ^{2} A =\cos ^{4} A$$\cos ^{2} A+\cos ^{4} A=1$

Given:\( \triangle \mathrm{ABC} \) is right angled at \( \mathrm{C} \)To do:We have to find the value of \( \cos (\mathrm{A}+\mathrm{B}) \).Solution:  \( \triangle \mathrm{ABC} \) is right angled at \( \mathrm{C} \)This implies,$\angle C=90^{\circ}$ We know that,In $\triangle A B C$,Sum of the three angles $=180^{\circ}$$\angle A+\angle B+\angle C=180^{\circ}$$\angle A+\angle B+90^{\circ} =180^{\circ}$$A+B=180^{\circ}-90^{\circ}$$A+B=90^{\circ}$Therefore,$\cos (A+B)=\cos 90^{\circ}=0$