Question

Sin(50+theta)-cos (40-theta)+tan1°tan10°tan20°tan70°tan80°tan89°

Answer 1

Here is the answer...

Answer 2

Answer:

$Value \: of \\sin(50+\theta)-cos(40-\theta)\\+tan1\degree tan10\degree tan20\degree tan70\degree tan80\degree tan89 \degree=1$

Step-by-step explanation:

$Value \: of \\sin(50+\theta)-cos(40-\theta)\\+tan1\degree tan10\degree tan20\degree tan70\degree tan80\degree tan89 \degree$

$=sin[90-(40-\theta)]-cos(40-\theta)\\+(tan1\degree tan89\degree) (tan10\degree tan80\degree) (tan20\degree tan70 \degree)$

$=cos(40-\theta)-cos(40-\theta)\\+[tan(90-89\degree)tan89] [tan(90-80\degree) tan80\degree] [tan(90-70\degree )tan70 \degree]$

$=0+(cot89tan89)(cot80tan80)(cot70tan70)$

$=1\times 1 \times 1$

$=1$

Therefore,

$Value \: of \\sin(50+\theta)-cos(40-\theta)\\+tan1\degree tan10\degree tan20\degree tan70\degree tan80\degree tan89 \degree=1$

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