Aspire Faculty ID #16654 · Topic: CUET 2023 · Just now
CUET 2023

If $ \cot^2 45^\circ - \sin^2 45^\circ = K \sin^2 30^\circ \times \tan^2 45^\circ \times \sec^2 45^\circ $, then the value of $K$ is

Solution

$ \cot 45^\circ = 1 \Rightarrow \cot^2 45^\circ = 1 $ $ \sin 45^\circ = \dfrac{1}{\sqrt{2}} \Rightarrow \sin^2 45^\circ = \dfrac{1}{2} $ LHS: $ \cot^2 45^\circ - \sin^2 45^\circ = 1 - \dfrac{1}{2} = \dfrac{1}{2} $ Now, $ \sin 30^\circ = \dfrac{1}{2} \Rightarrow \sin^2 30^\circ = \dfrac{1}{4} $ $ \tan 45^\circ = 1 \Rightarrow \tan^2 45^\circ = 1 $ $ \sec 45^\circ = \sqrt{2} \Rightarrow \sec^2 45^\circ = 2 $ RHS: $ K \times \dfrac{1}{4} \times 1 \times 2 = \dfrac{K}{2} $ Equating LHS and RHS: $ \dfrac{1}{2} = \dfrac{K}{2} \Rightarrow K = 1 $

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