1.

Without using trigonometric tables, prove that(i) \(\frac{2sin68^\circ}{cos22^\circ}-\frac{2cot15^\circ}{5tan75^\circ}- \frac{3tan45^\circ{tan20^\circ}tan40^\circ{tan50^\circ}tan70^\circ}5\) = 12sin68°/cos22°-2cot15°/5tan75°-3tan45°tan20°tan40°tan50°tan70°/5 = 1(ii) \(\frac{7cos55^\circ}{3sin35^\circ}-\frac{4(cos70^\circ{cosec20^\circ})}{3(tan5​​^\circ{tan25^\circ}tan45^\circ{tan65^\circ{tan85^\circ}})}\) = 17cos55°/3sin35° - 4(cos70°cosec20°)/3(tan5°tan45°tan65°tan85°)

Answer»

(i) LHS = \(\frac{2sin68^\circ}{cos22^\circ}-\frac{2cot15^\circ}{5tan75^\circ}- \frac{3{tan45^\circ}{tan20^\circ}{tan40^\circ}{tan50^\circ}{tan70^\circ}}5\)

\(\frac{2sin68^\circ}{sin(90^\circ-22^\circ)}-\frac{2cot15^\circ}{5tan75^\circ}- \frac{3\times1\times{cot(90^\circ-20^\circ)}\times{cot(90^\circ-40^\circ)}\times{tan50^\circ}\times{tan70^\circ}}5\)

\(\frac{2sin68^\circ}{sin68^\circ}-\frac{2cot15^\circ}{5cot15^\circ}- \frac{3\times{cot70^\circ}{tan50^\circ}tan70^\circ}5\)

= 2 - \(\frac{2}5\)\(\frac{3\times{\frac{1}{tan70^\circ}}\times{\frac{1}{tan50^\circ}}\times{tan50^\circ}\times{tan70^\circ}}5\)

= 2 - \(\frac{2}5\)\(\frac{3}5\)

\(\frac{10-2-3}5\)

\(\frac{5}5\)

= 1

= RHS

(ii) LHS =  \(\frac{7cos55^\circ}{3sin35^\circ}-\frac{4(cos70^\circ{cosec20^\circ})}{3(tan5​​°tan25^\circ{tan45^\circ}tan65^\circ{tan85^\circ})}\)

\(\frac{7cos55^\circ}{3cos(90^\circ-35^\circ)}-\frac{4(sin(90^\circ-70^\circ)cosec 20^\circ}{{3cot(90^\circ- 5^\circ)}\times{cot(90^\circ-25^\circ)}\times1\times{tan65^\circ}\times{tan85^\circ}}\)

\(\frac{7cos55^\circ}{3cos55^\circ}\) - \(\frac{{4(sin20^\circ}{cosec20^\circ})}{{3({cot85^\circ}{cot65^\circ}{tan65^\circ}{tan85^\circ}}}\)

\(\frac{7}3\) - \(\frac{{4(sin20^\circ}\times{\frac{1}{tan20^\circ}})}{{3({\frac{1}{tan85^\circ}}\times{\frac{1}{tan65^\circ}}{tan65^\circ}{tan85^\circ)}}}\) 

\(\frac{7}3\) - \(\frac{4}3\)

\(\frac{3}3\)

= 1

= RHS



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