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Question : The value of $\text{cot 17°}(\text{cot 73°} \cos^{2}22°+\frac{1}{\cot 17°\sec^{2}68°})$ is:

Option 1: $0$

Option 2: $1$

Option 3: $2$

Option 4: $\sqrt{3}$


Team Careers360 9th Jan, 2024
Answer (1)
Team Careers360 17th Jan, 2024

Correct Answer: $1$


Solution : Use: $\cot 73°= \tan(90°-73°)= \tan17°$ and $\cos^{2}22° = \cos^{2}(90°-68°) = \sin^2 68°$
$\text{cot 17°}(\text{cot 73°} \cos^{2}22°+\frac{1}{\cot17°\sec^{2}68°})$
= $\text{cot 17°}(\text{tan 17°} \sin^{2}68°+{\tan17°\cos^{2}68°})$
= $\text{cot 17°}[\text{tan 17°}(\sin^{2}68°+\cos^{2}68°)]$
= $\text{cot 17°}\text{tan 17°}$ [$\because(\sin^{2}68°+\cos^{2}68°)=1]$
= $\text{cot 17°}\frac{1}{\cot 17°}$
= $1$
Hence, the correct answer is 1.

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