Question

Pls solvethis question..​

Answer 1

Answer:

2

Step-by-step explanation:

$\huge\mathcal{Given}\\(cos^225°+cos^265°)+cosec\theta. sec(90-\theta)- cot\theta. Tan(90-\theta)\\=(cos^225°+cos^2(90-25°)+cosec\theta. sec(90-\theta)- cot\theta. Tan(90-\theta)\\\bold{we\:\:know\:\:that}\\\bold{cos(90-x)=sinx}\\\bold{sec(90-x)=cosec x}\\\bold{tan(90-x)=cotx}\\=(cos^225°+sin^225°)+cosec\theta. cosec(\theta)- cot\theta. cot\theta\\=(cos^225°+sin^225°)+(cosec^2\theta- cot^2\theta)\\\bold{we\:\:know\:\:that}\\\bold{sin^2x+cos^2x =1}\\\bold{cosec^2x-cot^2x=1}\\=(cos^225°+sin^225°)+(cosec^2\theta- cot^2\theta)\\= 1+1 =2.$

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Answer 2

refer to the attachment...

hope it helps ya

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