Question

2log cos theata +log(1+
tan²) theta

Answer 1

$\implies2 log( \cos\theta ) + log(1 + {tan}^{2}\theta )$

We know that :- 1 + tan²θ = sec²θ

$\implies2 log( \cos\theta ) + log( \sec ^{2}\theta )$

Now , a log(b ) = log( b^a)

$\implies log ({\cos^{2}\theta}) + log( \sec ^{2}\theta )$

Now , log a + log b = log ab

$\implies log ({\cos^{2}\theta} \times \sec ^{2}\theta)$

we know that secθ = 1/ cos θ

$\implies log ({\cos^{2}\theta} \times \dfrac{1}{\cos^{2}\theta} )$

$\implies log1$

we know log1 = 0

$\implies0$

Answer : 0

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$\large\bf\blue{Additional-Information}$

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

Answer 2

Answer:

0

Step-by-step explanation:

We know that :- 1 + tan²θ = sec²θ

Now , a log(b ) = log( b^a)

Now , log a + log b = log ab

we know that secθ = 1/ cos θ

we know log1 = 0

Answer : 0