All Exams Test series for 1 year @ ₹349 only
Question

Which of the following is a FALSE statement?

This question was previously asked in
SSC CGL 2023 (Tier-II) Paper 1 Previous Year Paper (26-Oct-2023) (Shift-1)
The correct answer is

tan2 A = 1 - sec2 A

Finding the False Trigonometry Statement

The question asks us to identify the statement that is FALSE among the given trigonometric identities. To do this, we will examine each statement and check if it holds true based on fundamental trigonometric identities.

Recalling Key Trigonometric Identities

We will use the following well-known trigonometric identities:

  • Pythagorean Identity: $\sin^2 A + \cos^2 A = 1$
  • Derived Pythagorean Identities:
    • $1 + \tan^2 A = \sec^2 A$
    • $1 + \cot^2 A = \cosec^2 A$
  • Reciprocal Identities:
    • $\sec A = \frac{1}{\cos A}$
    • $\cosec A = \frac{1}{\sin A}$
    • $\cot A = \frac{1}{\tan A}$
  • Ratio Identities:
    • $\tan A = \frac{\sin A}{\cos A}$
    • $\cot A = \frac{\cos A}{\sin A}$

Analyzing Each Trigonometric Statement

Let's check each given statement one by one:

Statement 1: $\tan^2 A = 1 - \sec^2 A$

We know the Pythagorean identity $1 + \tan^2 A = \sec^2 A$. Let's rearrange this identity to express $\tan^2 A$:

From $1 + \tan^2 A = \sec^2 A$, we subtract 1 from both sides:

$\tan^2 A = \sec^2 A - 1$

Now let's look at the given statement: $\tan^2 A = 1 - \sec^2 A$. We can rewrite $1 - \sec^2 A$ as $-(\sec^2 A - 1)$.

So the statement is $\tan^2 A = -(\sec^2 A - 1)$.

Comparing this with the correct identity $\tan^2 A = \sec^2 A - 1$, we see that the given statement claims $\tan^2 A = - \tan^2 A$. This is only true if $\tan^2 A = 0$, which means $\tan A = 0$. This identity does not hold true for all values of A where $\tan A$ and $\sec A$ are defined. Therefore, this statement is FALSE.

Statement 2: $\sin A = \tan A \times \cos A$

We know the ratio identity $\tan A = \frac{\sin A}{\cos A}$ (assuming $\cos A \neq 0$). Let's substitute this into the right side of the statement:

Right Side = $\tan A \times \cos A = \left(\frac{\sin A}{\cos A}\right) \times \cos A$

Assuming $\cos A \neq 0$, we can cancel $\cos A$:

Right Side = $\sin A$

The statement becomes $\sin A = \sin A$, which is TRUE for all values of A where $\tan A$ and $\cos A$ are defined and $\cos A \neq 0$.

Statement 3: $\cosec^2 A - 1 = \cot^2 A$

We know the Pythagorean identity $1 + \cot^2 A = \cosec^2 A$. Let's rearrange this identity to isolate $\cot^2 A$:

From $1 + \cot^2 A = \cosec^2 A$, we subtract 1 from both sides:

$\cot^2 A = \cosec^2 A - 1$

This matches the given statement exactly. Therefore, this statement is TRUE for all values of A where $\cot A$ and $\cosec A$ are defined.

Statement 4: $\cos A \times \sec A = 1$

We know the reciprocal identity $\sec A = \frac{1}{\cos A}$ (assuming $\cos A \neq 0$). Let's substitute this into the left side of the statement:

Left Side = $\cos A \times \sec A = \cos A \times \left(\frac{1}{\cos A}\right)$

Assuming $\cos A \neq 0$, we can cancel $\cos A$:

Left Side = 1

The statement becomes $1 = 1$, which is TRUE for all values of A where $\cos A$ and $\sec A$ are defined and $\cos A \neq 0$.

Conclusion: Identifying the False Trigonometry Statement

Based on our analysis of each trigonometric statement, we found that statement 1 is the only one that is FALSE. The correct identity relating $\tan^2 A$ and $\sec^2 A$ is $\tan^2 A = \sec^2 A - 1$, not $\tan^2 A = 1 - \sec^2 A$.

Statement Analysis Truth Value
$\tan^2 A = 1 - \sec^2 A$ Derived from $1 + \tan^2 A = \sec^2 A$ gives $\tan^2 A = \sec^2 A - 1$. The given statement is $\tan^2 A = -(\sec^2 A - 1) = -\tan^2 A$, which is not generally true. FALSE
$\sin A = \tan A \times \cos A$ Substitute $\tan A = \frac{\sin A}{\cos A}$. RHS = $\frac{\sin A}{\cos A} \times \cos A = \sin A$ (for $\cos A \neq 0$). LHS = RHS. TRUE
$\cosec^2 A - 1 = \cot^2 A$ Rearranging $1 + \cot^2 A = \cosec^2 A$ gives $\cot^2 A = \cosec^2 A - 1$. Matches the statement. TRUE
$\cos A \times \sec A = 1$ Substitute $\sec A = \frac{1}{\cos A}$. LHS = $\cos A \times \frac{1}{\cos A} = 1$ (for $\cos A \neq 0$). LHS = RHS. TRUE

Revision Table: Fundamental Trigonometric Identities

Type of Identity Identity Notes
Pythagorean $\sin^2 A + \cos^2 A = 1$ Foundation identity
Derived Pythagorean $1 + \tan^2 A = \sec^2 A$ Derived by dividing $\sin^2 A + \cos^2 A = 1$ by $\cos^2 A$
Derived Pythagorean $1 + \cot^2 A = \cosec^2 A$ Derived by dividing $\sin^2 A + \cos^2 A = 1$ by $\sin^2 A$
Reciprocal $\sec A = \frac{1}{\cos A}$ Valid when $\cos A \neq 0$
Reciprocal $\cosec A = \frac{1}{\sin A}$ Valid when $\sin A \neq 0$
Reciprocal $\cot A = \frac{1}{\tan A}$ Valid when $\tan A \neq 0$
Ratio $\tan A = \frac{\sin A}{\cos A}$ Valid when $\cos A \neq 0$
Ratio $\cot A = \frac{\cos A}{\sin A}$ Valid when $\sin A \neq 0$

Additional Information on Trigonometric Identities

Trigonometric identities are equations that are true for all values of the variable for which the expressions are defined. They are crucial in simplifying trigonometric expressions, solving trigonometric equations, and calculating derivatives and integrals of trigonometric functions.

  • The domain of validity for trigonometric identities depends on the specific functions involved. For example, identities involving $\tan A$ or $\sec A$ are not defined when $\cos A = 0$ (i.e., $A = \frac{\pi}{2} + n\pi$ for integer $n$). Identities involving $\cot A$ or $\cosec A$ are not defined when $\sin A = 0$ (i.e., $A = n\pi$ for integer $n$).
  • Understanding the relationship between the six trigonometric functions (sin, cos, tan, cot, sec, cosec) is key to mastering identities. They are interconnected through reciprocal, ratio, and Pythagorean relationships.
  • Practicing with different types of problems, such as simplifying expressions or proving identities, helps reinforce the understanding and application of these fundamental trigonometric identities.
Was this answer helpful?

Similar Questions

  1. If a = 45° and b = 15°, what is the value of \({\cos (a - b ) - \cos (a + b)} \over {\cos(a - b) + \cos(a + b)}\)?

  2. What is the value of cosec 15° sec 15°?

  3. If cosec θ + cot θ = p, then the value of \({{p^2 \ - \ 1} \over p^2 \ + \ 1}\) is:

  4. Find the value of cos 2A cos 2B + sin2(A - B) - sin2(A + B)

  5. Find the value of tan 27° tan 34° + tan 34° tan 29° + tan 29° tan 27°.

  6. If sin2 θ − 3 sin θ + 2 = 0, then find the value of θ (0° ≤ θ ≤ 90°).

  7. What is the value of sin 75° + sin 15°?

  8. Find the value of secθ - tanθ, if secθ + tanθ = \(\sqrt5\).

  9. If sin (5x - 25°) = cos(5y + 25°), where 5x - 25° and 5y + 25° are acute angles,then the value of (x + y) is:

  10. If tan (A + B) = √3 and tan (A - B) = \(\frac{1}{\sqrt 3}\); 0° < (A + B) < 90°; A > B, then the values of A and B are _______ respectively.


Important Questions from Trigonometry

  1. (secθ + tanθ)/(secθ - tanθ)  is equal to:

  2. If tan 45°, cot θ then the value of θ, in radians is

  3. ABC is a triangle If sin (A+B)/2 = √3/2, then the value of sin C/2 is

  4. The angles of elevation of the top of a temple, from the foot and the top of a building 30 m high, are 60° and 30° respectively. Then height of the temple is

  5. what is the principal value of \(\sin^{-1} \left( \sin \dfrac{2 \pi}{3} \right)\)  ?

Need Expert Advice?
Upcoming Exams
SSC CGL
September 30, 2026
UPSSSC PET
October 23, 2026
Test Series
SSC CGL img
SSC
SSC CGL (Tier I + Tier II) 2026 Mock Test Series - Latest Pattern
2503 Tests 6 Tests Free
5314 Attempts
4.2(864)
English, Hindi

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App