Question

2 log (cos teta)+log(1+tan²teta)=​

Answer 1

Answer:

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Answer 2

$\tt \longrightarrow 2 \: log( cos \theta) + log(1 + {tan}^{2} \theta) \\ \\ \sf we \: know \: that : \rightarrow \boxed{ \tt \large1 + {tan}^{2} \theta= {sec}^{2} \theta } \\ \\ \tt \longrightarrow 2 \: log( cos \theta) + log( {sec}^{2} \theta)\\ \\ \tt \longrightarrow 2 \: log( cos \theta) + log( \frac{1}{{cos}^{2} \theta} )\\ \\ \tt \longrightarrow 2 \: log( cos \theta) + log{(cos \: \theta)}^{ - 2} \\ \\ \tt \longrightarrow 2 \: log( cos \theta) - 2 \: log{(cos \: \theta)} \\ \\ \tt \longrightarrow 0$

Answer : 0

Trigonometric Identities :

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$