CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Trigonometric Ratios of Compound Angles

Prove that : ...

Question

Prove that : $\frac{t a n θ + s e c θ - 1}{t a n θ - s e c θ + 1}$ = sec $θ$ + tan $θ$ = $\frac{1 + s i n θ}{c o s θ}$ .

Open in App

Solution

LHS = $\frac{t a n θ + s e c θ - 1}{t a n θ - s e c θ + 1}$

LHS = $\frac{(t a n θ + s e c θ) - 1}{(t a n θ - s e c θ) + 1}$

LHS = $\frac{(s e c θ + t a n θ) - (s e c^{2} θ - t a n^{2} θ)}{t a n θ - s e c θ + 1}$

LHS = $\frac{(s e c θ + t a n θ) - (s e c θ + t a n θ) (s e c θ - t a n θ)}{t a n θ - s e c θ + 1}$

LHS = $\frac{(s e c θ + t a n θ) [1 - (s e c θ - t a n θ)]}{t a n θ - s e c θ + 1}$

LHS = $\frac{(s e c θ + t a n θ) (1 - s e c θ + t a n θ)}{t a n θ - s e c θ + 1}$

LHS = $\frac{(s e c θ + t a n θ) (t a n θ - s e c θ + 1)}{(t a n θ - s e c θ + 1)}$

LHS = $s e c θ + t a n θ$

LHS = $\frac{1}{c o s θ} + \frac{s i n θ}{c o s θ}$ = $\frac{1 + s i n θ}{c o s θ}$ = RHS

Suggest Corrections

thumbs-up

1

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Compound Angles

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Trigonometric Ratios of Compound Angles

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon