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.

​

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

Answer:

★ 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°}}}

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:

Step-by-step explanation:

${}$