Question

log tan 10° + log tan 80º =
A) O
B) 1
HASHYAM'S​

Answer 1

$\displaystyle \sf{log(tan {10}^{ \circ}) + log(tan {80}^{ \circ}) } = \bf \: 0$

Given :

$\displaystyle \sf{log(tan {10}^{ \circ}) + log(tan {80}^{ \circ}) }$

To find :

The value of the expression is

A) 0

B) 1

Solution :

Step 1 of 2 :

Write down the given expression

Here the given expression is

$\displaystyle \sf{log(tan {10}^{ \circ}) + log(tan {80}^{ \circ}) }$

Step 2 of 2 :

Simplify the given expression

$\displaystyle \sf{log(tan {10}^{ \circ}) + log(tan {80}^{ \circ}) }$

$\displaystyle \sf{ = log(tan {10}^{ \circ}) + log \bigg(tan( {90}^{ \circ} - {10}^{ \circ})\bigg) }$

$\displaystyle \sf{ =log(tan \: {10}^{ \circ}) + log(cot \: {10}^{ \circ}) }$

$\displaystyle \sf{ = log(tan {30}^{ \circ} \times cot {10}^{ \circ}) \: \: \: \bigg[ \: \because \:log \: a + log \: b = log \: (ab) \bigg] }$

$\displaystyle \sf{ =log \: \bigg( tan \: {10}^{ \circ} \times \frac{1}{tan \: {10}^{ \circ}} \bigg) }$

$\displaystyle \sf{ = log\: 1 }$

$= 0$

Hence the correct option is A) 0

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Answer 2

Answer:

The value of  (log tan10° + log tan80°) will be equal to zero.

Step-by-step explanation:

We know that $logA+logB=logAB$

The given expression:

$log (tan 10^{o}) + log (tan 80^{o})$

        $= log (tan 10^{o}) + log (tan (90^{o}-10^{o}))\\ = log (tan 10^{o}) + log (cot10^{o})$            ∵  $[tan(90^{o}-\theta )=cot\theta]$

        $=log (tan10^{o}*cot10^{o})$                                 ∵ $[tan\theta=\frac{1}{cot\theta}]$

        $=log(tan10^{o}*\frac{1}{tan10^{o}}) \\ =log1\\ =0$

Therefore the value of the given expression is equal to zero. So the option A is correct.