If tanθ + secθ = l, then prove that secθ = l² + 1/2l

Solution:

Given, tanθ + secθ = l ---------------------- (1)

We have to prove that secθ = l² + 1/2l

Considering LHS: tanθ + secθ

Multiplying and dividing by (secθ - tanθ),

(tanθ + secθ)(secθ - tanθ)/(secθ - tanθ)

By using algebraic identity,

(a² - b²) = (a - b)(a + b)

So, (tanθ + secθ)(secθ - tanθ) = (sec²θ - tan²θ)

Now LHS = (sec²θ - tan²θ) / (secθ - tanθ)

By using trigonometric identity,

1 + tan² A = sec² A

On rearranging,

sec² A - tan² A = 1

So, (sec²θ - tan²θ) / (secθ - tanθ) = 1/(secθ - tanθ)

Now, 1/(secθ - tanθ) = l

(secθ - tanθ) = 1/l ---------------------- (2)

Adding (1) and (2),

(tanθ + secθ) + (secθ - tanθ) = l + 1/l

2secθ = l + 1/l

On simplification,

2secθ = (l² + 1)/l

secθ = (l² + 1)/2l

Therefore, secθ = l² + 1/2l

✦ Try This: If secθ = x, write the value of tanθ.

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 8

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NCERT Exemplar Class 10 Maths Exercise 8.4 Problem 9

If tanθ + secθ = l, then prove that secθ = l² + 1/2l

Summary:

The tangent function can be expressed as the ratio of the sine function and cosine function. If tanθ + secθ = l, it is proven that secθ = l²+1/2l

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☛ Related Questions:

• If sinθ + cosθ = p and secθ + cosecθ = q, then prove that q (p2 – 1) = 2p
• If a sinθ + b cosθ = c, then prove that a cosθ – b sinθ = √(a2+b2-c2)
• Prove that (1+secθ-tanθ)/(1+secθ+tanθ) = (1-sinθ)/cosθ