sec²A • cosec ²A = tan² A+ cot ²A +2
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here is your answer,
we use identity,
sec²A- tan²A=1
cosec²A-cot²A=1
so the answer is,
sec²A×cosec²A=tan²A+cot²A+2
(1+tan²A)(1+cot²A)=tan²A+cot²A+2
by multiplying,we get
1+tan²A+cot²A+1=tan²A+cot²A+2
tan²A+cot²A+2=tan²A+cot²A+2
LHS=RHS
proved
we use identity,
sec²A- tan²A=1
cosec²A-cot²A=1
so the answer is,
sec²A×cosec²A=tan²A+cot²A+2
(1+tan²A)(1+cot²A)=tan²A+cot²A+2
by multiplying,we get
1+tan²A+cot²A+1=tan²A+cot²A+2
tan²A+cot²A+2=tan²A+cot²A+2
LHS=RHS
proved
harsh5359:
I will solve it
expand sin 5x in power of sin x.
I didn't know about this because I am study in class 9 so calcus is not in my syllabus
Bhai its not calculs
so what is it
de moivre's theorem's and application
bsc math trigonometry ka hai bhai
I didn't know because it is not included in my syllabus of 9 so I don't know
ok
thanks Bhai ye question dene ke liye mein esko try karunga
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