Question

Evaluate: $\sin ({50}^o+\theta )-\cos ({40}^o-\theta )+\tan 1^o\tan {10}^o\tan {20}^o\tan {70}^o\tan {80}^o\tan {89}^o$

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B $1$

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

$=\sin [{90}^o-{40}^o+\theta ]-\cos ({40}^o-\theta )+\left(\tan 1^o\tan {89}^o\right)\left(\tan {10}^o\tan {80}^o\right)\left(\tan {20}^o\tan {70}^o\right)$
$=\sin [{90}^o-({40}^o-\theta \left)\right]-\cos ({40}^o-\theta )+\tan ({90}^o-{89}^o)\tan {89}^o\tan {(90}^o-{80}^o)\tan {80}^o\tan {(90}^o-{70}^o)\tan {70}^o$
$=\cos ({40}^o-\theta )-\cos ({40}^o-\theta )+\cot {89}^o\tan {89}^o\cot {80}^o\tan {80}^o\cot {70}^o\tan {70}^o$
$=0+1$ $\because \cot \theta \tan \theta =1$]
$=1$

Hence, $\sin ({50}^o+\theta )-\cos ({40}^o-\theta )+\tan 1^o\tan {10}^o\tan {20}^o\tan {70}^o\tan {80}^o\tan {89}^o=1$