135⁰
135⁰ = 135\(\times\) \(\frac { \pi }{ 180 } =\frac { 3\pi }{ 4 } \)

30⁰
30⁰ = 30 \(\times\) \(\frac { \pi }{ 180 } =\frac { \pi }{ 6 } \)

Let sec-1\(\left( -\sqrt { 2 } \right) \) = y
⇒  -\(\sqrt { 2 } \) = sec y
⇒ sec y = -sec\(\frac { \pi }{ 4 } \)
⇒ sec y = sec\(\left( \pi -\frac { \pi }{ 4 } \right) \)      [∵ sec is negative in the II quad]
⇒ y = \(\frac { 3\pi }{ 4 } \)
Thus, the principal of sec-1(\(\sqrt { 2 } \)) is \(\frac { 3\pi }{ 4 } \) .

Let cosec-1(-1) = y, where \(-\frac { \pi }{ 2 } \le y\le \frac { \pi }{ 2 } \)
⇒ -1 = cosec y
⇒ cosec y = -cosec \(\left( \frac { \pi }{ 2 } \right) \)
⇒ cosec y = cosec -\(\left( \frac { \pi }{ 2 } \right) \)    [∵ cosec (-ፀ) = -cosec ፀ]
⇒ y = -\(\left( \frac { \pi }{ 2 } \right) \)
Thus, the principal value of cosec-1 is -\(\left( \frac { \pi }{ 2 } \right) \).

Let \(sin^{ -1 }\left( \frac { 1 }{ \sqrt { 2 } } \right) \) = y, where \(-\frac { \pi }{ 2 } \le y\le \frac { \pi }{ 2 } \)
⇒ sin y = \(\frac { 1 }{ \sqrt { 2 } } \)
⇒ sin y = sin \(\frac { \pi }{ 4 } \)
⇒ y = \(\frac { \pi }{ 4 } \)
Thus the principal value of \(sin^{ -1 }\left( \frac { 1 }{ \sqrt { 2 } } \right) \)= \(\frac { \pi }{ 4 } \) .

\(\cos { \left( \pi +\theta \right) } =-\cos { \theta } \)
LHS = \(\cos { \left( \pi +\theta \right) } =-\cos { \theta } \)
= RHS
Hence proved.

sin 105^o = sin (60 + 45)
= sin 60 cos 45 + cos 60 sin 45
= \(\frac { \sqrt { 3 } }{ 2 } .\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ 2 } .\frac { 1 }{ \sqrt { 2 } } =\frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } \)

Sin 25⁰  is an acute angle, 25⁰ lies in the I quadrant

Given tan²θ = 1- k²
Adding 1 both sides we get,
1 + tan²θ = 1 + 1-k²
⇒ sec²θ = 2-k²
Taking power 3/2 both sides we get,
(sec²θ)^3/2 = (2-k²)^3/2
⇒ sec³θ = (2-k²)^3/2
⇒ sec θ sec²θ = (2-k²)^3/2
⇒ sec θ + sec θ tan²θ = (2-k²)^3/2
⇒ sec θ +\(\frac{1}{cos\theta}.\frac{sin^2\theta}{cos^2\theta}\) = (2-k²)^3/2
⇒ sec θ + tan θ.cosec θ.tan²θ = (2-k²)^3/2
⇒ sec \(\theta\) + cosec \(\theta\) tan³ \(\theta\)  = (2- k²)^3/2
Hence proved.