Prove the following :

(i) We have to prove: $\sin \theta \cdot \sin \left(90^{\circ}-\theta\right)-\cos \theta \cdot \cos \left(90^{\circ}-\theta\right)=0$

Left hand side

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

$=\sin \theta \cdot \cos \theta-\cos \theta \cdot \sin \theta$

$=\sin \theta(\cos \theta-\cos \theta)$

$=0$

=Right hand side

Proved

(ii) We have to prove: $\frac{\cos \left(90^{\circ}-\theta\right) \cdot \sec \left(90^{-}-\theta\right) \cdot \tan \theta}{\operatorname{cosec}\left(90^{-}-\theta\right) \cdot \sin \left(90^{\circ}-\theta\right) \cdot \cot \left(90^{\circ}-\theta\right)}+\frac{\tan \left(90^{\circ}-\theta\right)}{\cot \theta}=2$

Left hand side

$=\frac{\cos \left(90^{\circ}-\theta\right) \cdot \sec \left(90^{\circ}-\theta\right) \cdot \tan \theta}{\operatorname{cosec}\left(90^{-}-\theta\right) \cdot \sin \left(90^{-}-\theta\right) \cdot \cot \left(90^{\circ}-\theta\right)}+\frac{\tan \left(90^{\circ}-\theta\right)}{\cot \theta}$

$=\frac{\sin \theta \cdot \operatorname{cosec} \theta \cdot \tan \theta}{\sec \theta \cdot \cos \theta \cdot \tan \theta}+\frac{\cot \theta}{\cot \theta}$

$=\frac{\tan \theta}{\tan \theta}+\frac{\cot \theta}{\cot \theta}$

$=1+1$

$=2$

= right hand side

Proved

(iii) We have to prove: $\frac{\tan \left(90^{\circ}-\mathrm{A}\right) \cot \mathrm{A}}{\operatorname{cosec}^{2} \mathrm{~A}}-\cos ^{2} A=0$

Left hand side

$=\frac{\tan \left(90^{\circ}-A\right) \cot A}{\operatorname{cosec}^{2} A}-\cos ^{2} A$

$=\frac{\cot A \cdot \cot A}{\operatorname{cosec}^{2} A}-\cos ^{2} A$

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

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

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

$=0$

= right hand side

Proved

(iv) We have to prove: $\frac{\cos \left(90^{\circ}-A\right) \cdot \sin \left(90^{\circ}-A\right)}{\tan \left(90^{\circ}-A\right)}=\sin ^{2} A$

Left hand side

$=\frac{\cos \left(90^{\circ}-A\right) \cdot \sin \left(90^{\circ}-A\right)}{\tan \left(90^{\circ}-A\right)}$

$=\frac{\sin A \cdot \cos A}{\cot A}=\sin ^{2} A$

$=\frac{\sin A \cdot \cos A \cdot \sin A}{\cos A}$

$=\sin ^{2} A$

= Right hand side

Proved

(v) We have to prove: $\sin \left(50^{\circ}+\theta\right)-\cos \left(40^{\circ}-\theta\right)+\tan 1^{\circ} \tan 10^{\circ} \tan 70^{\prime \prime} \tan 80^{\circ} \tan 89=1$

Left hand side

$=\sin \left(50^{\circ}+\theta\right)-\cos \left(40^{\circ}-\theta\right)+\tan 1^{\circ} \tan 10^{\circ} \tan 20^{\circ} \tan 70^{\circ} \tan 80^{\circ} \tan 89^{\circ}$

$=\cos \left[90^{\circ}-\left(50^{\circ}+\theta\right)\right]-\cos \left(40^{\circ}-\theta\right)+\tan \left(90^{\circ}-89^{\circ}\right) \tan \left(90^{\circ}-80^{\circ}\right) \tan \left(90^{\circ}-70^{\circ}\right) \tan 70^{\circ} \tan 80^{\circ} \tan 89^{\circ}$

$=\cos \left(40^{\circ}-\theta\right)-\cos \left(40^{\circ}-\theta\right)+\cot 89^{\circ} \cdot \cot 80^{\circ} \cdot \cot 70^{\circ} \cdot \tan 70^{\circ} \cdot \tan 80^{\circ} \cdot \tan 89^{\circ}$

$=0+\left(\tan 89^{\circ} \cdot \cot 89^{\circ}\right)\left(\tan 80^{\circ} \cdot \cot 80^{\circ}\right)\left(\tan 70^{\circ} \cdot \cot 70^{\circ}\right)$

Since $\tan \theta \cdot \cot \theta=1$. So

$=1 \times 1 \times 1$

$=1$

=Right hand side

Proved