Question

Prove that:

(i) $sin\theta cos({90}^{\circ }-\theta )+sin({90}^{\circ }-\theta )cos\theta =1$

(ii) $\frac{sin\theta }{cos({90}^{\circ }-\theta )}+\frac{cos\theta }{sin({90}^{\circ }-\theta )}=2$

(iii) $\frac{sin\theta cos({90}^{\circ }-\theta )cos\theta }{sin({90}^{\circ }-\theta )}+\frac{cos\theta sin({90}^{\circ }-\theta )sin\theta }{cos({90}^{\circ }-\theta )}=1$

(iv) $\frac{cos({90}^{\circ }-\theta )sec({90}^{\circ }-\theta )tan\theta }{cosec({90}^{\circ }-\theta )sin({90}^{\circ }-\theta )cot({90}^{\circ }-\theta )}+\frac{tan({90}^{\circ }-\theta )}{cot\theta }=2$

(v) $\frac{cos({90}^{\circ }-\theta )}{1+sin({90}^{\circ }-\theta )}+\frac{1+sin({90}^{\circ }-\theta )}{cos({90}^{\circ }-\theta )}=2cosec\theta$

(vi) $\frac{sec({90}^{\circ }-\theta )cosec\theta -tan({90}^{\circ }-\theta )cot\theta +co{s}^2{25}^{\circ }+co{s}^2{65}^{\circ }}{3tan{27}^{\circ }tan{63}^{\circ }}=\frac{2}{3}$

(vii) $cot\theta tan({90}^{\circ }-\theta )-sec({90}^{\circ }-\theta )cosec\theta +\sqrt{3}tan{12}^{\circ }tan{60}^{\circ }tan{78}^{\circ }=2$

Answer 1

(i) $sin\theta cos({90}^{\circ }-\theta )+sin({90}^{\circ }-\theta )cos\theta =1LHS=sin\theta cos({90}^{\circ }-\theta )+sin({90}^{\circ }-\theta )cos\theta =sin\theta sin\theta +cos\theta cos\theta =si{n}^2\theta +co{s}^2\theta =1=RHS$

Hence Proved

(ii) $\frac{sin\theta }{cos({90}^{\circ }-\theta )}+\frac{cos\theta }{sin({90}^{\circ }-\theta )}=2LHS=\frac{sin\theta }{cos({90}^{\circ }-\theta )}+\frac{cos\theta }{sin({90}^{\circ }-\theta )}=\frac{sin\theta }{sin\theta }+\frac{cos\theta }{cos\theta }=1+1=2=RHS$

Hence Proved

(iii) $\frac{sin\theta cos({90}^{\circ }-\theta )cos\theta }{sin({90}^{\circ }-\theta )}+\frac{cos\theta sin({90}^{\circ }-\theta )sin\theta }{cos({90}^{\circ }-\theta )}=1LHS=\frac{sin\theta cos({90}^{\circ }-\theta )cos\theta }{sin({90}^{\circ }-\theta )}+\frac{cos\theta sin({90}^{\circ }-\theta )sin\theta }{cos({90}^{\circ }-\theta )}=\frac{sin\theta sin\theta cos\theta }{cos\left(\theta \right)}+\frac{cos\theta cos\theta sin\theta }{sin\theta }=si{n}^2\theta +co{s}^2\theta =1=RHS$

Hence Proved

(iv) $\frac{cos({90}^{\circ }-\theta )sec({90}^{\circ }-\theta )tan\theta }{cosec({90}^{\circ }-\theta )sin({90}^{\circ }-\theta )cot({90}^{\circ }-\theta )}+\frac{tan({90}^{\circ }-\theta )}{cot\theta }=2LHS=\frac{cos({90}^{\circ }-\theta )sec({90}^{\circ }-\theta )tan\theta }{cosec({90}^{\circ }-\theta )sin({90}^{\circ }-\theta )cot({90}^{\circ }-\theta )}+\frac{tan({90}^{\circ }-\theta )}{cot\theta }=\frac{sin\theta cosec\theta tan\theta }{sec\theta cos\theta tan\theta }+\frac{cot\theta }{cot\theta }=1+1=2=RHS$

Hence Proved

(v) $\frac{cos({90}^{\circ }-\theta )}{1+sin({90}^{\circ }-\theta )}+\frac{1+sin({90}^{\circ }-\theta )}{cos({90}^{\circ }-\theta )}=2cosec\theta LHS=\frac{cos({90}^{\circ }-\theta )}{1+sin({90}^{\circ }-\theta )}+\frac{1+sin({90}^{\circ }-\theta )}{cos({90}^{\circ }-\theta )}=\frac{sin\theta }{1+cos\theta }+\frac{1+cos\theta }{sin\theta }=\frac{si{n}^2\theta +(1+cos\theta {)}^2}{(1+cos\theta )sin\theta }=\frac{si{n}^2\theta +1+co{s}^2\theta +2cos\theta }{(1+cos\theta )sin\theta }=\frac{1+1+2cos\theta }{(1+cos\theta )sin\theta }=\frac{2+2cos\theta }{(1+cos\theta )sin\theta }=\frac{2(1+cos\theta )}{(1+cos\theta )sin\theta }=\frac{2}{sin\theta }=2cosec\theta =RHS$

Hence Proved

(vi) $\frac{sec({90}^{\circ }-\theta )cosec\theta -tan({90}^{\circ }-\theta )cot\theta +co{s}^2{25}^{\circ }+co{s}^2{65}^{\circ }}{3tan{27}^{\circ }tan{63}^{\circ }}=\frac{2}{3}LHS=\frac{sec({90}^{\circ }-\theta )cosec\theta -tan({90}^{\circ }-\theta )cot\theta +co{s}^2{25}^{\circ }+co{s}^2{65}^{\circ }}{3tan{27}^{\circ }tan{63}^{\circ }}=\frac{cosec\theta cosec\theta –cot\theta cot\theta +si{n}^2({90}^{\circ }–{25}^{\circ })+co{s}^2{65}^{\circ }}{3tan{27}^{\circ }cot({90}^{\circ }–{63}^{\circ })}=\frac{cose{c}^2\Theta –co{t}^2+si{n}^2{65}^{\circ }+co{s}^2{65}^{\circ }}{3tan{27}^{\circ }cot{27}^{\circ }}=\frac{1+1}{3}=\frac{2}{3}=RHS$

(vii) $cot\theta tan({90}^{\circ }-\theta )-sec({90}^{\circ }-\theta )cosec\theta +\sqrt{3}tan{12}^{\circ }tan{60}^{\circ }tan{78}^{\circ }=2LHS=cot\theta tan({90}^{\circ }-\theta )-sec({90}^{\circ }-\theta )cosec\theta +\sqrt{3}tan{12}^{\circ }tan{60}^{\circ }tan{78}^{\circ }=cot\theta cot\theta –cosec\theta cosec\theta +\sqrt{3}tan{12}^{\circ }\times \sqrt{3}\times cot({90}^{\circ }–{78}^{\circ })=co{t}^2\theta –cose{c}^2\theta +3tan{12}^{\circ }cot{12}^{\circ }=-1+3=2=RHS$