Step-by-step explanation:
<h3><u>Given :-</u></h3>
[1+(1/Tan²θ)] + [ 1+(1/Cot²θ)]
<h3>
<u>Required To Prove :-</u></h3>
[1+(1/Tan²θ)]+[1+(1/Cot²θ)] = 1/(Sin²θ-Sin⁴θ)
<h3><u>Proof :-</u></h3>
On taking LHS
[1+(1/Tan²θ)] + [ 1+(1/Cot²θ)]
We know that
Tan θ = 1/ Cot θ
and
Cot θ = 1/Tan θ
=> (1+Cot²θ)(1+Tan²θ)
=> (Cosec² θ) (Sec²θ)
Since Cosec²θ - Cot²θ = 1 and
Sec²θ - Tan²θ = 1
=> (1/Sin² θ)(1/Cos² θ)
Since , Cosec θ = 1/Sinθ
and Sec θ = 1/Cosθ
=> 1/(Sin²θ Cos²θ)
We know that Sin²θ+Cos²θ = 1
=> 1/[(Sin²θ)(1-Sin²θ)]
=> 1/(Sin²θ-Sin²θ Sin²θ)
=> 1/(Sin²θ - Sin⁴θ)
=> RHS
=> LHS = RHS
<u>Hence, Proved.</u>
<h3><u>Answer:-</u></h3>
[1+(1/Tan²θ)]+[1+(1/Cot²θ)] = 1/(Sin²θ-Sin⁴θ)
<h3><u>Used formulae:-</u></h3>
→ Tan θ = 1/ Cot θ
→ Cot θ = 1/Tan θ
→ Cosec θ = 1/Sinθ
→ Sec θ = 1/Cosθ
<h3><u>Used Identities :-</u></h3>
→ Cosec²θ - Cot²θ = 1
→ Sec²θ - Tan²θ = 1
→ Sin²θ+Cos²θ = 1
Hope this helps!!
Answer:
∠ACD= ∠BAC ( reason: CPCTC)
<h3 /><h3>
</h3>
Equation for Perimeter of a rectangle: Perimeter = 2W + 2L
<h3>Defining the variables, let</h3>
<h3>Width = x</h3><h3>Length = 2x+3 (3 more than twice the width)</h3>
<h3>Plugging everything into the equation</h3>
<h3>30= 2(x) + 2(2x+3) using the distributive property,</h3>
<h3>30=2x+4x+6 combining like terms</h3>
<h3>30=6x+6 subtracting 6 from both sides,</h3>
<h3>24=6x divide both sides by 6</h3>
<h3>4=x This means that the width is 4 m.</h3>
<h3>To get the length, use the expression L=2x+3 and plug in x = 4 that was already solved for</h3>
<h3>L=2(4)+3</h3>
<h3>L=8+3 = 11 m</h3>
<h3>So the dimensions of the rectangle are width is 4 m and length is 11 m.</h3>
Is there any more to this question? Because-5 doesn’t belong anywhere
