## Write 'True' or 'False' and justify your answer in each of the following:The value of the expression $\left(\cos ^{2} 23^{\circ}-\sin ^{2} 67^{\circ}\right)$ is positive.

Updated on 10-Oct-2022 13:28:58

## $\sin \left(45^{\circ}+\theta\right)-\cos \left(45^{\circ}-\theta\right)$ is equal to(A) $2 \cos \theta$(B) 0(C) $2 \sin \theta$(D) 1

Updated on 10-Oct-2022 13:28:58
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$

## If $\sin \theta-\cos \theta=0$, then the value of $\left(\sin ^{4} \theta+\cos ^{4} \theta\right)$ is(A) 1(B) $\frac{3}{4}$(C) $\frac{1}{2}$(D) $\frac{1}{4}$

Updated on 10-Oct-2022 13:28:58
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}$$\tan 45^{\circ}=1Therefore,\sin \theta-\cos \theta=0$$\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}$$=\frac{1}{2} ## If $4 \tan \theta=3$, then $\left(\frac{4 \sin \theta-\cos \theta}{4 \sin \theta+\cos \theta}\right)$ is equal to(A) $\frac{2}{3}$(B) $\frac{1}{3}$(C) $\frac{1}{2}$(D) $\frac{3}{4}$ Updated on 10-Oct-2022 13:28:58 Given:$4 \tan \theta=3$To do:We have to find the value of $\left(\frac{4 \sin \theta-\cos \theta}{4 \sin \theta+\cos \theta}\right)$.Solution:4 \tan \theta=3$$\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}$$=\frac{1}{2} ## The value of the expression $\left[\frac{\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}}{\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}\right]$ is(A) 3(B) 2(C) 1(D) 0 Updated on 10-Oct-2022 13:28:58 Given:$\left[\frac{\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}}{\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}\right]$To do:We have to find the value of the expression $\left[\frac{\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}}{\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}\right]$.Solution:We know that, \sin \left(90^{\circ}-\theta\right)=\cos \theta$$\cos \left(90^{\circ}-\theta\right)=\sin \theta$$\sin ^{2} \theta+\cos ^{2} \theta=1Therefore, [\frac{\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}}{\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}]=\frac{\sin ^{2} 22^{\circ}+\sin ^{2}(90^{\circ}-22^{\circ})}{\cos ^{2}(90^{\circ}-68^{\circ})+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin (90^{\circ}-63^{\circ})$$=\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$$=2Read More ## Given that $\sin \alpha=\frac{1}{2}$ and $\cos \beta=\frac{1}{2}$, then the value of $(\alpha+\beta)$ is(A) $0^{\circ}$(B) $30^{\circ}$(C) $60^{\circ}$(D) $90^{\circ}$ Updated on 10-Oct-2022 13:28:52 Given:$\sin \alpha=\frac{1}{2}$ and $\cos \beta=\frac{1}{2}$To do:We have to find the value of $(\alpha+\beta)$.Solution: \sin \alpha =\frac{1}{2}$$=\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}$$=90^{\circ}The value of $(\alpha+\beta)$ is 90^{\circ}. ## If $\sin \mathrm{A}+\sin ^{2} \mathrm{~A}=1$, then the value of the expression $\left(\cos ^{2} \mathrm{~A}+\cos ^{4} \mathrm{~A}\right)$ is(A) 1(B) $\frac{1}{2}$(C) 2(D) 3 Updated on 10-Oct-2022 13:28:52 Given:$\sin \mathrm{A}+\sin ^{2} \mathrm{~A}=1$To do:We have to find the value of the expression $\left(\cos ^{2} \mathrm{~A}+\cos ^{4} \mathrm{~A}\right)$.Solution: We know that,\sin ^{2} A+\cos ^{2} A=1Therefore,\sin A+\sin ^{2} A =1$$\sin A =1-\sin ^{2} A$$=\cos ^{2} ASquaring on both sides, we get,\sin ^{2} A=\cos ^{4} A$$1-\cos ^{2} A =\cos ^{4} A$$\cos ^{2} A+\cos ^{4} A=1 ## If $\triangle \mathrm{ABC}$ is right angled at $\mathrm{C}$, then the value of $\cos (\mathrm{A}+\mathrm{B})$ is(A) 0(B) 1(C) $\frac{1}{2}$(D) $\frac{\sqrt{3}}{2}$ Updated on 10-Oct-2022 13:28:52 Given:$\triangle \mathrm{ABC}$ is right angled at $\mathrm{C}$To do:We have to find the value of $\cos (\mathrm{A}+\mathrm{B})$.Solution: $\triangle \mathrm{ABC}$ is right angled at $\mathrm{C}$This implies,\angle C=90^{\circ} We know that,In \triangle A B C,Sum of the three angles =180^{\circ}$$\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$
Previous 1 ... 4 5 6 7 8 ... 775 Next