What is the maximum value of 1 sec ⁡ θ ?

The maximum value of 1 sec ⁡ θ is 1 because the maximum value of cos ⁡ θ is 1 . (Since, 1 sec ⁡ θ = cos ⁡ θ )

What is the maximum value of 1 cosec ⁡ θ ?

We know that: 1 cosec ⁡ θ = sin ⁡ θ The maximum value of 1 cosec ⁡ θ is 1 , because the maximum value of sin ⁡ θ is 1 .

If tan ⁡ θ = 4 5 , find the value of cos ⁡ θ - sin ⁡ θ cos ⁡ θ + sin ⁡ θ .

Given that tan ⁡ θ = 4 5 Now, cos ⁡ θ - sin ⁡ θ cos ⁡ θ + sin ⁡ θ = 1 - sin ⁡ θ cos ⁡ θ 1 + sin ⁡ θ cos ⁡ θ [Dividing both numerator and denominator by cos ⁡ θ ] = 1 - tan ⁡ θ 1 + tan ⁡ θ = 1 - 4 5 1 + 4 5 = 1 9

If cos ⁡ θ = 2 3 , find the value of sec ⁡ θ - 1 sec ⁡ θ + 1 .

Given that, cos ⁡ θ = 2 3 We have to find sec ⁡ θ - 1 sec ⁡ θ + 1 ⇒ sec ⁡ θ - 1 sec ⁡ θ + 1 = 3 2 - 1 3 2 + 1 ( ∵ cos x = 1 sec x ) ⇒ sec ⁡ θ - 1 sec ⁡ θ + 1 = 1 2 5 2 ⇒ sec ⁡ θ - 1 sec ⁡ θ + 1 = 1 5 Hence, the value of given expression is 1 5 .

If 3 cot ⁡ θ = 4 , find the value of 4 cos ⁡ θ - sin ⁡ θ 2 cos ⁡ θ + sin ⁡ θ .

We have: cot ⁡ θ = 4 3 We know that in a right angle triangle, 4 cos ⁡ θ - sin ⁡ θ 2 cos ⁡ θ + sin ⁡ θ = 4 cotθ - 1 2 cotθ + 1 ( Dividing by sinθ ) = 4 × 4 3 - 1 2 × 4 3 + 1 = 16 - 3 8 + 3 = 13 11 Hence, the value of 4 cos ⁡ θ - sin ⁡ θ 2 cos ⁡ θ + sin ⁡ θ is 13 11 .

Given tan ⁡ θ = 1 5 , what is the value of c o s e c 2 θ - sec 2 ⁡ θ c o s e c 2 θ + sec 2 ⁡ θ = ?

Given: tan ⁡ θ = 1 5 We know that: tan ⁡ θ = P e r p e n d i c u l a r B a s e Applying Pythagoras theorem, H y p o t e n u s e = P e r p e n d i c u l a r 2 + B a s e 2 ⇒ H y p o t e n u s e = 5 x 2 + x 2 = 6 x Now, cosec 2 θ - sec 2 ⁡ θ cosec 2 θ + sec 2 ⁡ θ = 6 2 1 2 - 6 2 5 2 6 2 1 2 + 6 2 5 2 ∵ cosecθ = Hypotenuse Opposite side ; secθ = Hypotenuse Adjacent side = 6 1 - 6 5 6 1 + 6 5 = 24 5 36 5 = 2 3 Hence, the value of c o s e c 2 θ - sec 2 ⁡ θ c o s e c 2 θ + sec 2 ⁡ θ is 2 3 .

If cot ⁡ θ = 1 3 , write the value of 1 - cos 2 ⁡ θ 2 - sin 2 ⁡ θ .

Given: cot ⁡ θ = 1 3 Let us assume that the Base of the triangle = x and Perpendicular = 3 x Applying Pythagoras Theorem, H y p o t e n u s e = P e r p e n d i c u l a r 2 + B a s e 2 ⇒ H y p o t e n u s e = 2 x Now we find, 1 - cos 2 ⁡ θ 2 - sin 2 ⁡ θ = 1 - 1 2 2 2 2 - 3 2 2 2 = 1 - 1 4 2 - 3 4 = 3 5 Hence, the value of 1 - cos 2 ⁡ θ 2 - sin 2 ⁡ θ is 3 5 .