If $x = \sec\phi - \tan\phi$ and $y = \csc\phi + \cot\phi$ then show that $xy + x - y + 1 = 0$.

If x = secφ - tanφ and y = cscφ + cotφ then show t – Class 11 Mathematics

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x = \sec\phi - \tan\phi

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Concept understanding — Trigonometric Identity Proof

Trigonometric Identity Proof: From Intuition to Precision

Imagine you're standing at the corner of a right triangle. The two shorter sides — one horizontal, one vertical — and the sloping hypotenuse are all connected. If you change the angle at your corner, the lengths of the sides change, but the relationship between them stays fixed. That fixed relationship is what a trigonometric identity captures.

The Core Idea

A trigonometric identity is an equation involving trigonometric functions (like $\sin \theta$ , $\cos \theta$ , $\tan \theta$ ) that is true for every angle $\theta$ where both sides are defined. It's not a conditional equation (like $\sin \theta = 0.5$ , which is true only for specific angles). It's an eternal truth about how these functions relate.

The most famous one is:

\sin^2 \theta + \cos^2 \theta = 1

This holds for any angle $\theta$ — acute, obtuse, negative, whatever. Why? Because on the unit circle, $\sin \theta$ is the $y$ -coordinate and $\cos \theta$ is the $x$ -coordinate of a point on a circle of radius 1. The Pythagorean theorem says $x^2 + y^2 = 1$ , so $\sin^2 \theta + \cos^2 \theta = 1$ is just the Pythagorean theorem in disguise.

Proving an Identity: The Method

When you're asked to prove a trigonometric identity, you're not solving for an angle. You're showing that the left-hand side (LHS) and right-hand side (RHS) are the same expression, just written differently.

The golden rule: Start with one side and transform it into the other, using known identities and algebraic manipulation. Never move terms across the equals sign as if solving an equation — that assumes the identity is already true, which is what you're trying to prove.

A Simple Example

Prove: $\tan \theta \cdot \cos \theta = \sin \theta$

Step 1: Pick a side to start with. Usually, the more complicated side is easier to simplify. Here, the LHS looks more complex.

Step 2: Replace $\tan \theta$ with $\frac{\sin \theta}{\cos \theta}$ (a known identity).

\tan \theta \cdot \cos \theta = \frac{\sin \theta}{\cos \theta} \cdot \cos \theta

Step 3: Cancel $\cos \theta$ (provided $\cos \theta \neq 0$ — but the identity holds for all angles where both sides are defined, and at $\cos \theta = 0$ , $\tan \theta$ is undefined anyway).

= \sin \theta

That's it. The LHS simplifies exactly to the RHS.

The Toolbox of Known Identities

To prove any identity, you need to know the basic building blocks:

Identity	Formula
Pythagorean	$\sin^2 \theta + \cos^2 \theta = 1$
Quotient	$\tan \theta = \frac{\sin \theta}{\cos \theta}$ , $\cot \theta = \frac{\cos \theta}{\sin \theta}$
Reciprocal	$\csc \theta = \frac{1}{\sin \theta}$ , $\sec \theta = \frac{1}{\cos \theta}$ , $\cot \theta = \frac{1}{\tan \theta}$
Even-Odd	$\sin(-\theta) = -\sin \theta$ , $\cos(-\theta) = \cos \theta$

Watch out

A common mistake is to treat $\sin^2 \theta$ as $(\sin \theta)^2$ — which it is — but then incorrectly think $\sin^2 \theta + \cos^2 \theta = 1$ means $\sin \theta + \cos \theta = 1$ . It does not. The square applies to the whole sine value, not to the angle.

A Slightly Harder Proof

Prove: $\frac{1 - \cos^2 \theta}{\cos \theta} = \sin \theta \tan \theta$

Start with LHS: $\frac{1 - \cos^2 \theta}{\cos \theta}$

From the Pythagorean identity, $1 - \cos^2 \theta = \sin^2 \theta$ . So:

\frac{\sin^2 \theta}{\cos \theta} = \sin \theta \cdot \frac{\sin \theta}{\cos \theta} = \sin \theta \tan \theta

That's the RHS. Done. …

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We simplify the given expressions for $x$ and $y$ by converting them to terms of $\sin\phi$ and $\cos\phi$ . Substituting these simplified forms into the expression $xy + x - y + 1$ and combining terms over a common denominator reveals that the numerator simplifies to $1 - (\sin^2\phi + \cos^2\phi)$ , which equals $0$ , thus proving the identity.

This problem asks us to prove a trigonometric identity involving expressions for $x$ and $y$ in terms of $\sec\phi$ , $\tan\phi$ , $\csc\phi$ , and $\cot\phi$ . The most straightforward approach for such problems is to convert all trigonometric functions into their fundamental forms, $\sin\phi$ and $\cos\phi$ . This often simplifies the expressions significantly, making algebraic manipulation easier. Once $x$ and $y$ are in terms of $\sin\phi$ and $\cos\phi$ , we can substitute them into the target expression $xy + x - y + 1$ and simplify it to show that it equals zero.

Let's proceed step-by-step:

Simplify the expression for $x$ .

The given expression for $x$ is $x = \sec\phi - \tan\phi$ .

We know the reciprocal identity $\sec\phi = \frac{1}{\cos\phi}$ and the quotient identity $\tan\phi = \frac{\sin\phi}{\cos\phi}$ .

Substituting these into the expression for $x$ :

$x = \frac{1}{\cos\phi} - \frac{\sin\phi}{\cos\phi}$

Since both terms have a common denominator, we can combine them:

$x = \frac{1 - \sin\phi}{\cos\phi}$
Simplify the expression for $y$ .

The given expression for $y$ is $y = \csc\phi + \cot\phi$ .

We know the reciprocal identity $\csc\phi = \frac{1}{\sin\phi}$ and the quotient identity $\cot\phi = \frac{\cos\phi}{\sin\phi}$ .

Substituting these into the expression for $y$ :

$y = \frac{1}{\sin\phi} + \frac{\cos\phi}{\sin\phi}$

Combining these terms over the common denominator:

$y = \frac{1 + \cos\phi}{\sin\phi}$
Substitute the simplified $x$ and $y$ into the target expression.

We need to show that $xy + x - y + 1 = 0$ . Let's substitute our simplified forms of $x$ and $y$ into the left-hand side (LHS) of this equation:

LHS $= \left(\frac{1 - \sin\phi}{\cos\phi}\right) \left(\frac{1 + \cos\phi}{\sin\phi}\right) + \left(\frac{1 - \sin\phi}{\cos\phi}\right) - \left(\frac{1 + \cos\phi}{\sin\phi}\right) + 1$
Combine the terms using a common denominator.

The common denominator for all terms will be $\sin\phi\cos\phi$ .

Let's expand the product term first:

$xy = \frac{(1 - \sin\phi)(1 + \cos\phi)}{\sin\phi\cos\phi} = \frac{1 + \cos\phi - \sin\phi - \sin\phi\cos\phi}{\sin\phi\cos\phi}$

Now, rewrite the entire LHS with the common denominator: …

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Q.If x=sec⁡ϕ−tan⁡ϕx = \sec\phi - \tan\phix=secϕ−tanϕ and y=csc⁡ϕ+cot⁡ϕy = \csc\phi + \cot\phiy=cscϕ+cotϕ then show that xy+x−y+1=0xy + x - y + 1 = 0xy+x−y+1=0.

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Q.If $x = \sec\phi - \tan\phi$ and $y = \csc\phi + \cot\phi$ then show that $xy + x - y + 1 = 0$ .

Unlock everything free for 14 days