1+tan^2(A) = sec^2(A) [Pythagorean Identities]
tan^2(A)cot(A) = tan(A)[tan(A)cot(A)] = tan(A)[1] = tan(A)
*see photo for complete solution*
<span>(1 + cos² 3θ) / (sin² 3θ) = 2 csc² 3θ - 1
Starting with the left: Note that cos²θ + </span><span>sin²θ = 1.
In the same way: </span><span>cos²3θ + <span>sin²3θ = 1
</span></span>Therefore cos²3θ = 1 - <span>sin²3θ
</span> From the top: (1 + cos² 3θ) = 1 + 1 - sin²3θ = 2 - <span>sin²3θ
</span>
(1 + cos² 3θ) / (sin² 3θ) = (<span>2 - sin²3θ) / (sin² 3θ) = 2/</span><span>sin² 3θ - </span><span>sin²3θ/</span>sin²3θ
= 2/<span>sin² 3θ - 1; But 1/</span><span>sinθ = csc</span><span>θ, Similarly </span>1/sin3θ = csc3θ
= 2 *(1/sin<span>3θ)² - 1</span>
= 2csc²3θ - 1. Therefore LHS = RHS. QED.
Add like terms
I'm assuming that you forgot to put the parenthasees at the end
so exg
4x+8x=12x
2x^2+3x^2=5x^2
3x^2+3x^3=3x^2+3x^3
group like tems
6x^3+9x-8+5x-9x^2+7
6x^3-9x^2+14x-1
Answer:
D. ⅔
Step-by-step explanation:
I hope it helps :)
