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Find the value of 2sec2A+6csc2A−2tan2A−6cot2A.
2
6
4
8
8
Group the terms with same coefficients: 2(sec²A - tan²A) + 6(cosec²A - cot²A). Using standard identities (sec²A - tan²A = 1) and (cosec²A - cot²A = 1), the expression becomes 2(1) + 6(1).
An expression 2sec²A + 6cosec²A - 2tan²A - 6cot²A.
1+tan2A=sec2A and 1+cot2A=csc2A
Group the terms with same coefficients: 2(sec²A - tan²A) + 6(cosec²A - cot²A). Using standard identities (sec²A - tan²A = 1) and (cosec²A - cot²A = 1), the expression becomes 2(1) + 6(1).
Many students try to convert everything to sin and cos, which leads to a long and complex calculation instead of using Pythagorean identities.
Group the terms
Regroup the terms having common coefficients to simplify the expression: 2(sec2A−tan2A)+6(csc2A−cot2A)
2(sec2A−tan2A)+6(csc2A−cot2A)
Apply trigonometric identities
Use the fundamental identities sec2A−tan2A=1 and csc2A−cot2A=1
2(1)+6(1)
Final calculation
Perform the final addition: 2+6=8
8
D is correct because the expression simplifies to 2(1) + 6(1) = 8.
These Pythagorean identities are frequently tested in exams; remember that sec2θ−tan2θ=1 and csc2θ−cot2θ=1 are the keys to solving complex-looking expressions.
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