Question

1)If sec 3 theta = √2, Find tan(90° - 2 theta)
2)The sums of squares of two consecutive positive odd numbers is 290. Find the number.


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Answer 1

$\huge\bold{\underline{Question\:1:}}$

If sec 3 theta = √2, Find tan(90° - 2 theta)

$\huge\bold{\underline{Answer:}}$

$\green\bigstar$ Given:

sec 3 theta = √2

$\blue\bigstar$ To find:

Find tan(90° - 2 theta)

$\purple\bigstar$ Solution:

⟹ sec 3θ = √2

⟹ sec 3θ = sec 45° [sec 45° = √2]

⟹ 3θ = 45°

⟹ $\boxed{\bf{\pink{θ = 15°}}}$

Now,

tan(90° - 2θ)

= cot 2θ

= cot 2 × 15° [ θ = 15°]

= cot 30°

= √3

Therefore, the value of tan(90°-2θ) = √3

$\red\bigstar$ Additional information:

T-RATIOS:

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3} }{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }& 1 & \sqrt{3} & \rm Not \: De fined \\ \\ \rm cosec A & \rm Not \: De fined & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm Not \: De fined \\ \\ \rm cot A & \rm Not \: De fined & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}\end{gathered}$

___________________________

$\huge\bold{\underline{Question\:2:}}$

The sums of squares of two consecutive positive odd numbers is 290. Find the number.

$\huge\bold{\underline{Answer:}}$

$\green\bigstar$ Given:

• The sums of squares of two consecutive positive odd numbers is 290

$\blue\bigstar$ To find:

Find the number.

$\purple\bigstar$ Solution:

Let the two consecutive number be x and (x+2)

According to question, we have

$\sf{:\implies x²+(x+2)²=290}$

$\sf{:\implies 2x²+4x+4=290}$

$\sf{:\implies 2x²+4x=290-4}$

$\sf{:\implies 2x²+4x=286}$

$\sf{:\implies 2(x²+2x)=286}$

$\sf{:\implies x²+2x=143}$

$\sf{:\implies x²+2x-143=0}$

$\sf{:\implies x²+13x-11x-143=0}$

$\sf{:\implies x(x+13)-11(x+13)=0}$

$\sf{:\implies (x+13)(x-11)=0}$

$\sf{:\implies x=-13}$

or,

$\boxed{\bf{\pink{⟹\:x=11}}}$

Since the number is positive,the number is x = 11

.°. x + 2 = 11 + 2 = 13

Therefore,

sum of numbers is = ( 11 + 13 ) = 24

Therefore, the number is 24

__________________________

Answer 2

Answers:-

1) Given:

sec 3θ = √2

√2 can be written as sec 45°.

So, sec 3θ = sec 45°

On comparing both sides we get,

⟶ 3θ = 45°

⟶ θ = 45/3

⟶ θ = 15°

Now,

We have to find;

⟶ tan (90 - 2θ)°

⟶ tan (90 - 2*15)°

⟶ tan (90 - 30)°

⟶ tan 60°

• tan 60° = √3

⟶ √3

∴ The value of tan (90 - 2θ)° is √3

___________________________

2) Given:

Sum of squares of two consecutive positive odd numbers = 290

Let the numbers be x , x + 2.

According to the above condition,

⟶ x² + (x + 2)² = 290

• (a + b)² = a² + b² + 2ab

⟶ x² + x² + 4 + 4x = 290

⟶ 2x² + 4x + 4 - 290 = 0

⟶ 2x² + 4x - 286 = 0

⟶ x² + 2x - 143 = 0

⟶ x² - 11x + 13x - 143 = 0

⟶ x(x - 11) + 13(x - 11) = 0

⟶ (x - 11)(x + 13) = 0

★ x - 11 = 0

⟶ x = 11

★ x + 13 = 0

⟶ x = - 13

It is given that they are positive integers. So, negative value of x is neglected.

∴

• First number = x = 11

• Second number = x + 2 = 11 + 2 = 13