2 log (cos teeta) + log (1+ tan² teeta) =
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Answer:
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Step-by-step explanation:
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⟹2log(cosθ)+log(1+tan
2
θ)
We know that :- 1 + tan²θ = sec²θ
\implies2 log( \cos\theta ) + log( \sec ^{2}\theta )⟹2log(cosθ)+log(sec
2
θ)
Now , a log(b ) = log( b^a)
\implies log ({\cos^{2}\theta}) + log( \sec ^{2}\theta )⟹log(cos
2
θ)+log(sec
2
θ)
Now , log a + log b = log ab
\implies log ({\cos^{2}\theta} \times \sec ^{2}\theta)⟹log(cos
2
θ×sec
2
θ)
we know that secθ = 1/ cos θ
\implies log ({\cos^{2}\theta} \times \dfrac{1}{\cos^{2}\theta} )⟹log(cos
2
θ×
cos
2
θ
1
)
\implies log1⟹log1
we know log1 = 0
\implies0⟹0
Answer : 0
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\large\bf\blue{Additional-Information}Additional−Information
1. Cosθ = base / hypotenuse
2. cossecθ = 1/ sinθ
3. sec θ = 1/cosθ
4. Cotθ = 1/ tanθ
5. Sin²θ+ Cos²θ= 1
6. Sec²θ - tan²θ = 1
7. cosec ²θ - cot²θ = 1
8. sin(90°−θ) = cos θ
9. cos(90°−θ) = sin θ
10. tan(90°−θ) = cot θ
11. cot(90°−θ) = tan θ
12. sec(90°−θ) = cosec θ
13. cosec(90°−θ) = sec θ
14. Sin2θ = 2 sinθ cosθ
15. cos2θ = Cos²θ- Sin²θ
16. 1 - sin² θ = cos²θ
17. 1 - cos²θ = sin²θ
18. 1 + tan² θ= sec²θ
20. cotA = 1/tanA
21. 1 + tan²A = sec²A
22. sec²A - tan²A = 1
23. sec²A - 1 = tan²A
24. 1 + cot²A = cosec²A
25. cosec²A - 1 = cot²A
26. cosec²A - cotA = 1
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